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Quadratic Equation: Solving by the square root method
This method can be used if the quadratic equation can be
put in the form au2 + bu + c = 0, where b = 0 and u is
an algebraic expression. In other words, there is no "bu"
term.
Example 1: Solve 3x2 + 36 = 0.
First, isolate x2.
3x2 = - 36
x2 = - 12
Next, take the square root of each side
and place a "" symbol in front of the root
on the right side.
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x 
2
 12
x    12
Quadratic Equation: Solving by the square root method
Last, simplify the radical expression.
x    12
x   12i
x   2 3i
Try to solve: 2x2 – 80 = 0.
The solutions set is: {  2 10 }.
Notes: The "" symbol is placed in front of the right side
because x 2  x . For example, x2 = 9 becomes
x 
2
9 or x 
9 which has two solutions,  3.
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Quadratic Equation: Solving by the square root method
Example 2: Solve (3x – 5)2 + 36 = 0.
Note: this has the form au2 + bu + c = 0, where b
= 0 and u is an algebraic expression (u = 3x – 5).
So first, isolate u2 = (3x – 5)2.
Next, take the square root of each side
and place a "" symbol in front of the root
(3x – 5)2 = - 36
3x  5   36
2
3x  5    36
on the right side.
Next, solve for x.
3x  5   36
5   36
x 
3
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Quadratic Equation: Solving by the square root method
Last, simplify the radical expression. If a fraction results,
simplify it. If nonreal solutions result it is customary to
write them in standard form (a + bi).
5   36
x 
,
3
5
6
x 
 i,
3
3
5  36i
x 
,
3
5
x 
 2i
3
5  6i
x 
,
3
Try to solve: Solve (4x + 1)2 = 24.
 1  2 6 
The solutions set is: 
.
4


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Quadratic Equation: Solving by the square root method
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