## KG1234 Group Title solve for x: (4x-3)^2=-15 one year ago one year ago

1. NickR

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

2. KG1234

yes

3. NickR

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. KG1234

16x^2-24x+24=0

6. NickR

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. NickR

9. NickR

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

10. KG1234

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

11. NickR

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

12. KG1234

okay yeah i did that

13. NickR

now find a common denominator

14. KG1234

8?

15. NickR

yup

16. KG1234

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

17. KG1234

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. NickR

yes

20. KG1234

then what do i do?

21. KG1234

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

22. NickR

i would do the quadratic formula

23. KG1234

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

24. Hero

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

25. NickR

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

26. NickR

yup

27. KG1234

i got it thanks to both of you!

28. NickR

not a problem

29. Hero

You got what? Mind sharing what you got?

30. KG1234

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

31. KG1234

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

32. NickR

yup