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Pythagorean Quadratic into binomials factoring quadratic equation.

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1 Answer

The 2 to the right of the first set of parenthesis looks like it might be an exponent because if it were multiply by 2 the 2 would usually be put to the left of the parenthesis.

So I'll do the problem two ways:

1.

(2x+6)^2 = (2x+9)^2
Take square root of both sides (N.B.: √(k^2) == |k| ):
|2x+6| = |2x+9|
2x+6 = ±(2x+9)

case 1: 2x+6 = +(2x+9)
6 = 9 is never true so there is no solution for this case.

case 2: 2x+6 = -(2x+9)
2x + 6 = -2x - 9
4x = -15
x = -15/4

check: (2(-15/4)+ 6)^2 =? (2(-15/4)+9)^2
(-15/2+12/2)^2 =? (-15/2+18/2)^2
(-3/2)^2 =? (3/2)^2
9/4 = 9/4 √ , so -15/4 is the only solution to problem 1.

2.

2*(2x+6) = (2x+9)^2
4x + 12 = 4x^2 + 36x + 81
0 = 4x^2 + 32x + 69

Use Quadratic Formula:
x = (-32 ± √(32^2-4(4)(69)))/(2(4))
x = -4 ± (√((32/4)^2-4(4)(69)/4^2))/2
x = -4 ± (√(8^2-69))/2
x = -4 ± (√(64-69))/2
x = -4 ± (i√(5))/2

check:
2*(2(-4 ± (i√(5))/2) + 6) =? (2(-4 ± (i√(5))/2) + 9)^2
2*(-8 ± i√(5) + 6) =? (-8 ± i√(5) + 9)^2
2*(-2 ± i√(5)) =? (1 ± i√(5))^2
-4 ± 2i√(5) =? 1 ± 2i√(5) - 5
-4 ± 2i√(5) = -4 ± 2i√(5) √

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