## anonymous 3 years ago solve for x: (4x-3)^2=-15

1. anonymous

so (4x-3)^2=(4x-3)(4x-3) right

2. anonymous

yes

3. anonymous

so take that and foil it

4. Hero

(4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9

5. anonymous

16x^2-24x+24=0

6. anonymous

but its not = to 0 its =-15

7. Hero

(4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9

8. anonymous

9. anonymous

so now is there a common number that they are all divisible by?

10. anonymous

wait im at 16x^2-24x+9=-15 right?

11. anonymous

ya then add the 15 to get 16x^2-24x+24=0

12. anonymous

okay yeah i did that

13. anonymous

now find a common denominator

14. anonymous

8?

15. anonymous

yup

16. anonymous

so does that then turn into 2x^2-3x=3=0?

17. anonymous

2x^2-3x+3 sorry

18. Hero

16x^2-24x+9=-15 16x^2 - 24x + 24 = 0 8(2x^2 - 3x + 3) = 0 2x^2 - 3x + 3 = 0 x^2 - 3/2 + 3/2 = 0 x^2 - 3/2x = -3/2 x^2 -3/2x + 9/16 = -3/2 + 9/16 (x - 3/4)^2 = -15/16 Solution is imaginary

19. anonymous

yes

20. anonymous

then what do i do?

21. anonymous

the answer is 3 + or - the square root of 15 divided by4 but idk how to get it

22. anonymous

i would do the quadratic formula

23. anonymous

-b + or - the square root of b^2 -4ac divided by 2a?

24. Hero

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

25. anonymous

(-b+-sqrt(b^2-4ac))/2a

26. anonymous

yup

27. anonymous

i got it thanks to both of you!

28. anonymous

not a problem

29. Hero

You got what? Mind sharing what you got?

30. anonymous

3 + or - the square root of 15 divided by 4

31. anonymous

3+ or - i square root 15 divided by 4 sorry i forgot the "i"

32. anonymous

yup