SOLUTION: I am having trouble reducing the fraction within the quadratic equation. Would you please check my process to see if I did it right and help me with the ending part? (4x-3)^2=0

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am having trouble reducing the fraction within the quadratic equation. Would you please check my process to see if I did it right and help me with the ending part? (4x-3)^2=0      Log On


   

Question 411024: I am having trouble reducing the fraction within the quadratic equation. Would you please check my process to see if I did it right and help me with the ending part?
(4x-3)^2=0
(4x-3)(4x-3)=0
16x^2-12x-12x-9=0
16x^2-24x-9=0
-24+- sqrt((24)^2-4(16)(-9))/(2(16))
-24+- sqrt(576+576)/(32)
-24+- sqrt(1152)/(32)
-24+- sqrt(8(144))/(32)
-24+- sqrt(4(2)(144))/(32)
-24+- 12(2) sqrt(2)/(32)
-24+- 24 sqrt(2)/(32)
-1+- sqrt(3(1)/(3)

Found 2 solutions by edjones, ewatrrr:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
16x^2-24x+9=0 not -9
x=(24+-sqrt(576-4*16*9))/32 not -24
=(24+-sqrt(0))/32
=24/32
=3/4
.
Ed

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

IF: 16x^2-24x-9=0  

THEN: 

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%2824+%2B-+sqrt%28+1152+%29%29%2F%2832%29+

[24 ± 24sqrt(2)]/(32)  = 3/4 ± (3/4)sqrt(2)



However, in this question: (4x-3)(4x-3)=0  | 16x^2-24x+9=0 

x+=+%2824+%2B-+sqrt%28+0+%29%29%2F%2832%29+

 x = 3/4