A quadratic equation is always written in the form of:
ax2 + bx + c = 0
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where a ≠ 0
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The
form
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ax2 + bx + c = 0 is called the standard
form of a
quadratic equation.
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Examples:
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x2
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− 5x + 6 = 0
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This is
a quadratic equation written in standard form.
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x2
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+ 4x = −4
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This is a quadratic equation that is not written in
standard form but
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can be
once we set the equation to: x2 + 4x + 4 = 0.
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x2
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= x
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This
too can be a quadratic equation once it is set to 0.
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x2 − x = 0 (standard form with c=0).
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Solving Quadratic Equations by Square Root Property
When x2 = a , where a is a real number, then your x
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= ±
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a
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Examples:
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x2 − 9 = 0
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y2 + 3 = 28
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x2 − 9 = 0
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y2 + 3 − 3 = 28 − 3
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x2
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= 9
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y2 = 25
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x
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= ±
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9
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y = ±
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25
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x
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= ±3
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y = ±5
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It can
also be solved by factoring the equation. Remember to always check your
solutions. You can use direct substitution of the solutions in the equation to
see if the solutions satisfy the equation.
Examples:
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x2 − 5x + 6 = 0
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(x − 3)(x − 2) = 0
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← Factoring x
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x − 3 = 0
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x − 2 = 0
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← Set it equal to 0 and solve for x
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x = 3
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x = 2
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Now check
if, x = 3 and x = 2 are the solutions of x2 − 5x + 6 = 0
Check:
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32
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− 5(3) + 6 = 0
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22 − 5(2) + 6 = 0
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9 − 15 + 6 = 0
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4 − 10 + 6 = 0
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2x2 + 7x − 4 = 0
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(2x − 1)(x + 4) = 0
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2x − 1 = 0
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x + 4 = 0
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2x = 1
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x = −4
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x = 1
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2
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Another method of checking the solutions is by using one of the
following statements:
The sum
of the solutions = − b
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or
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The
product of the solutions =
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c
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a
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a
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ax2 + bx + c = 0 .
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where a, b,
and c are the coefficients in
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Now we
check if x = 1 and x =
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− 4 are the solutions of 2x2
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+ 7x − 4 = 0
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2
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Check:
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Using
the sum of the solutions =
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1
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+ (−4) = −
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7
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2
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2
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Based
on the original equation =
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−
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b
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= −
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7
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2
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a
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Now by
using the product of the solutions =
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1
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(−4) = −2
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2
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c
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=
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−4
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= −2
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a
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2
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1 + 2x − x82 = 0 ← Rewrite in standard form by multiplying each side of the equation by x2
x2 + 2x − 8 = 0
(x + 4)(x − 2) = 0
x + 4 = 0
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x − 2 = 0
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x = −4
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x = 2
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Check:
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1 +
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2
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−
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8
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= 0
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−4
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(−4)2
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1 −
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1
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−
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1
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= 0
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2
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2
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1 +
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2
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−
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8
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= 0
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2
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22
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1 + 1 – 2
= 0
ß Solutions must be checked in the original equation to avoid any errors.
ß ← The two rational solutions
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Solution Using the Quadratic Formula
Factoring is useful only for those quadratic equations which have whole
numbers. When you encounter quadratic equations that can not be easily factored
out, use the quadratic formula to find the value of x:
x =
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−b ±
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b2 − 4ac
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2a
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Examples:
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x2
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− 8 = −2x
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x2
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+ 2x − 8 = 0 ← Rewrite in standard form,
where a = 1, b = 2, and c = −8
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−2 ±
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x =
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4 − 4(1)(−8)
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← Plug in numbers into the
equation
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2(1)
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= −2
± 
36 2(1)
36 2(1)
= −2
± 6 2
= 2,−4 ← The two rational solutions
3x2 − 13x + 4 = 0
x = −(−13)
± 
(−13)2 − 4(3)(4)
2(3)
= 13 ± 
121 6
121 6
= 
13 ± 11
13 ± 11
6
= 246 , 62 = 4, 13
In some
cases you encounter repeated rational solutions. And to prove you have the
right values you use the discriminant which gives you information about the
nature of the solutions to the equation. Based on
the expression b2 − 4ac , which is under the radical in the
quadratic formula it can be found in the equation ax2 + bx + c = 0.
I. When the discriminant is equal
to 0, the equation has repeated rational solutions.
Example: x2 − 2x + 1 = 0
By using
the discriminant b2 − 4ac = (−2)2 − 4(1)(1) = 0
x = −(−2)
± 
(−2)2 − 4(1)(1) 2(1)
(−2)2 − 4(1)(1) 2(1)
= 2 ± 
0 2
0 2
x = 1,1 ← Repeated rational solutions
II.
When the discriminant is positive
and a perfect square, the equation has two distinct rational solutions.
Example: x2 − 4x + 3 = 0
By
discriminant b2 − 4ac = (−4)2 − 4(1)(3) = 4
x = −(−4)
± (−4)2 − 4(1)(3) 2(1)
(−4)2 − 4(1)(3) 2(1)
= 4 ± 4 2
4 2
x = 3,1 ← Two distinct rational solutions
III. When
the discriminant is positive but not a perfect square, the equation has two
irrational solutions.
Example:
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x2 + 4x − 6 = 0
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The discriminant
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b2 − 4ac = (4)2 − 4(1)(−6) = 40
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x
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=
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−4 ±
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(4)2 −
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4(1)(−6)
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2(1)
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−4 ±
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=
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40
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2
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x = −2 ±
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10
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← Two irrational solutions
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IV. When the discriminant is
negative, the equation has two complex number solutions.
Example:
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x2 + 4x + 6 = 0
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The discriminant
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b2 − 4ac = (4)2 − 4(1)(6) = −8
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x =
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−4 ±
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(4)2 −
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4(1)(6)
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2(1)
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−4 ±
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x =
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−8
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← Two complex number solutions
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2
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One more method of solving
quadratic equations is by completing the square.
Example: Solve x 2 + 6 x + 5 = 0 by
completing the square.
1)
If the leading coefficient is not
1, use the multiplication (or division) property of equality to make it 1:
x 2 + 6 x + 5 = 0 ← In this
case the leading coefficient is already 1
2) Rewrite
the equation by sending the constant to the right side of the equation:
x 2 + 6x + 5 = 0
x2 + 6x + 5 − 5 = 0 − 5 x2 + 6x = −5
3) Divide the numerical coefficient the middle term by 2, then square
it, and add it to both sides of the equation, but leave the square form on the
left side of the equation:
x 2
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+ 6 x = −5
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x 2 + 6 x + ( 3 )2
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= −5 + (3)2
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6
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= 3 → (3)2
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x 2 + 6 x + ( 3 )2
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= −5
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Middle term coefficient = 6
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+ 9
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2
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x 2 + 6 x + ( 3 )2
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= 4
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4) Once you found the squared
number rewrite the equation as follows:
x 2 + 6 x + ( 3 )2 = 4 ← Bring down the variable x and put it inside the parentheses
← Use the sign of the middle term. In this case it is +. (x + 3
)2 = 4
← Write the squared number. In this case it is 3.
The
resultant binomial is (x + 3 )2 = 4
5) Using the square root property
clear the term.
(x + 3 )2
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= ±
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← The square root of a squared
term is the term by itself.
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4
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x + 3 = ±2
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6)
Solve for the variable x.
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x + 3 = ±2
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← The ± notation
is used because the square root can have both
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positive
and negative answers.
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x + 3 = 2
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x + 3 = −2
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(−1 + 3 )2 = (2 )2 = 4 And (−5 + 3 )2 = (−2 )2 = 4
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x = 2 − 3
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x = −2 − 3
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x = −1
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x = −5
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x = −1
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x = −5
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x 2 + 6x + 5 = 0
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x 2 + 6x + 5 = 0
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(−1)2 + 6(−1) + 5 = 0
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← Both solutions are true:
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(−5 )2 + 6(−5 ) + 5 = 0
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1− 6 + 5 = 0
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25 − 30 + 5 = 0
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0 = 0
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0 = 0
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Let’s keep practicing with one
more.
Example:
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4 x 2 − 2 x − 5 = 0
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← One way to make the leading
coefficient 1 is
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14 (4 x 2 − 2 x − 5 ) = 14 (0)
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by
multiplying both sides of the equation by 14
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x 2 −
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1 x −
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5 = 0
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← Move the constant to the right side of the equation
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2
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4
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x 2 −
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1 x =
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5
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2
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4
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x
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2
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−
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1
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x +
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1
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2
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5
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1
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2
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← Divide the middle term
coefficient by 2, square it,
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2
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=
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4
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+
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4
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4
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and add
it to both sides of the equation:
1
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÷
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2
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=
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1
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×
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1
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1
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→
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1
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2
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|||||
2
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1
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2
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=
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4
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4
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||||||||||
2
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|||||||||||||||
2
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1
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1
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2
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5
|
1
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||||||||||
x
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−
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x +
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=
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+
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|||||||||||
2
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4
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4
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16
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x
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2
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−
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1
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1
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2
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=
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21
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2
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x +
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4
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16
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1
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2
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21
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← Write the squared number in
binomial form.
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x −
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4
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=
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16
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1
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2
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|||||||||||||
= ±
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21
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← Find the square root of both
sides and don’t forget the
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±
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x −
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4
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16
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||||||||||||
sign.
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||||||||||||||
1
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x −
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= ±
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21
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← Send the other number to the
right side of the
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|||||||||||
4
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||||||||||||||
16
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||||||||||||||
x − 14 = ± 
421
421
x = ± 421 + 14
421 + 14
x = 1 ± 21 4
21 4
equation.
← Try to solve it using the square root. If not possible leave it in
radical form.
← Solve for
x.
← Final
answer.
Solve each of the following equations by the method
of your choice and check your solutions.
1.
x2 − 2x + 1 = 0
2.
x2 + 9x + 20 = 0
3.
3x2 − 5x − 12 = 0
4.
6x2 + 9x − 6 = 0
5.
x2 + 3x − 28 = 0
6.
3x2 − 2x = 2x + 7
7.
4x2 − 12x = 16
8.
x2 + 3x = 0
9.
3 + 1x = 10x2
10.
3y2 − y − 4 = 0
11.
y2 + 2 y + 1 = 0
12.
x2 − 2x − 8 = 0
13.
x2 + 4 = 0
14.
x2 + x = −1
15.
9 y2 + 6y − 8 = 0
16.
y2 − 25 = 0
17.
6y2 − 13y + 6 = 0
18.
x − 34x = −31
19.
x − 4x = 215
20.
3 + 25x = x12
21.
4 − 1x = x32
22.
8x = x2
23.
(x − 5)(x + 8) =
−20
24.
(x + 6)(x − 3) = 10
25.
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2
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−
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x
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= 1
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x + 5
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x − 5
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1.
|
x2 − 2x + 1 = 0
|
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(x − 1)(x − 1) = 0
|
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x − 1 = 0
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x − 1 = 0
|
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x = 1
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x = 1
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3.
|
3x2 − 5x − 12 = 0
|
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(3x + 4)(x − 3) = 0
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3x + 4 = 0
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x − 3 = 0
|
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x = − 4
|
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x = 3
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3
|
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5.
|
x2 + 3x − 28 = 0
|
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(x + 7)(x − 4) = 0
|
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x + 7 = 0
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x − 4 = 0
|
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x = −7
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x = 4
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7. 4x2 − 12x = 16
4x2 − 12x − 16 = 0
4(x2 − 3x − 4) = 0
4(x − 4)(x + 1) = 0
x − 4 = 0 x + 1 = 0
x = 4 x = −1
2.
|
x2 + 9x + 20 = 0
|
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(x + 5)(x + 4) = 0
|
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x + 5 = 0
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x + 4 = 0
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x = −5
|
x = −4
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4. 6x2 + 9x − 6 = 0
|
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3(2x2 + 3x − 2) = 0
|
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3(2x − 1)(x + 2) = 0
|
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2x − 1 = 0
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x + 2 = 0
|
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x = 1
|
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x = −2
|
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2
|
|||
6. 3x2 − 2x = 2x + 7
|
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3x2 − 2x − 2x − 7 = 0
|
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3x2 − 4x − 7 = 0
|
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(3x − 7)(x + 1) = 0
|
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3x − 7 = 0
|
x + 1 = 0
|
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x = 7
|
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x = −1
|
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3
|
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8.
|
x2 + 3x = 0
|
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x(x + 3) = 0
|
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x = 0
|
x + 3 = 0
|
x = −3
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9.
|
3 +
|
1
|
=
|
10
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10. 3y2 − y − 4 = 0
|
||
x2
|
|||||||
x
|
|||||||
3x2 + x = 10
|
(3y − 4)( y + 1) = 0
|
||||||
3x2 + x − 10 = 0
|
3y − 4 = 0
|
||||||
y + 1 = 0
|
|||||||
(3x − 5)(x + 2) = 0
|
y = 4
|
||||||
y = −1
|
|||||||
3x − 5 = 0
|
3
|
||||||
x + 2 = 0
|
|||||||
x = 5
|
|||||||
x = −2
|
|||||||
3
|
|||||||
11. y2 + 2 y + 1 = 0 ( y + 1)2 = 0
y + 1 = 0 y = −1, −1
13. x2 + 4 = 0
x = −0
± 
0 − 4(1)(4) 2
0 − 4(1)(4) 2
x = ±
|
−16
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||
2
|
|||
15. 9 y2 + 6y − 8 = 0
x = −6
± 
36 − 4(9)(−8) 2(9)
36 − 4(9)(−8) 2(9)
=
|
−6 ±
|
324
|
|||
18
|
|||||
=
|
−6 ±18
|
||||
18
|
|||||
x =
|
6 +18
|
x =
|
6 −18
|
||
18
|
18
|
||||
x =
|
2
|
x =
|
−4
|
||
3
|
3
|
||||
17. 6y2 − 13y + 6 = 0
(3y − 2)(2 y − 3) = 0
3y − 2 = 0
|
2 y − 3 = 0
|
|||
y =
|
2
|
y =
|
3
|
|
3
|
2
|
|||
12. x2 − 2x − 8 = 0
(x − 4)(x + 2) = 0
x − 4 = 0 x + 2 = 0
x = 4 x = −2
14. x2 + x = −1
x 2 + x + 1 = 0
−1 ±
|
|||||
x =
|
1− 4(1)(1)
|
||||
2(1)
|
|||||
−1 ±
|
|||||
x =
|
−3
|
||||
2
|
|||||
16. y2 − 25 = 0
( y − 5)( y + 5) = 0
y − 5 = 0 y + 5 = 0
y = 5 y = −5
18. x − 34x = − 13
3x2 − 4 = − x
3x2 + x − 4 = 0
(3x + 4)(x − 1) = 0
3x + 4 = 0
|
x − 1 = 0
|
|
4
|
||
x = −
|
x = 1
|
|
3
|
||
19. x
−
|
4
|
=
|
21
|
|
x
|
||||
5
|
||||
5x2
|
− 20 = 21x
|
|||
5x2
|
− 21x − 20 = 0
|
|||
(5x + 4)(x − 5) = 0
|
||
5x + 4 = 0
|
x − 5 = 0
|
|
x = − 4
|
||
x = 5
|
||
5
|
21. 4 − 1x = x32
4x2 − x = 3
4x2 − x − 3 = 0
(4x + 3)(x − 1) = 0
4x + 3 = 0
|
x − 1 = 0
|
||
x = −
|
3
|
x = 1
|
|
4
|
|||
23. (x − 5)(x + 8) = −20
x2 + 3x − 40 + 20 = 0 x2 + 3x − 20 = 0
x = −3
± 
9 − 4(1)(−20) 2
9 − 4(1)(−20) 2
x = −3
± 
89 2
89 2
25.
x +2 5 − x −x 5 = 1
2(x − 5) − x(x + 5) = x2 − 25
2x − 10 − x2 − 5x − x2 + 25 = 0 −2x2
− 3x + 15
= 0
2x2 + 3x − 15 = 0
x =
|
−3 ±
|
9 − 4(2)(−15)
|
x = −3 ±
|
4
|
129
|
4
|
|||||
20. 
3 + 25x = x12
3 + 25x = x12
6x2 + 5x = 2
6x2 + 5x − 2 = 0
x = −5
± 
25 − 4(6)(−2)
25 − 4(6)(−2)
12
x = −5
± 73 12
73 12
22. 8x = x2
x 2 − 8x = 0
x(x − 8) = 0
x = 0x − 8 = 0
x
= 8
24. (x + 6)(x − 3) = 10
x2
+ 3x − 18
− 10 = 0
x2 + 3x − 28 = 0
(x + 7)(x − 4) = 0
x + 7 = 0 x − 4 = 0
x = −7 x = 4
