Write 3x² +8x-25 in the form a(x + h) + k. Where a, h and k are real numbers?

Question

Hence or otherwise, determine the minimum value of 3x² +8x-25

Solve the equation 3x² +8x-25=0

Answer 1

3(x² + 8/3 x - 25/3) = 0
x² + (8/3) x - 25/3 = 0
x² + (8/3) x + 16/9 = 25/3 + 16/9
(x + 4/3)² = 75/9 + 48/9
(x + 4/3)² = 123/9
(x + 4/3) = +/- (123)^(1/2) / 3
x = (- 4/3) +/-(123)^(1/2) / 3

Answer 2

I believe you want it in the form a(x+h)^2 + k by using completion of square method.
3x² +8x-25 = 3(x^2 + 8/3) - 25
= 3(x^2 +8/3 + 16/9) - 25 - 16/3
= 3(x + 4/3)^2 -91/3
a = 3, h = 4/3 and k = -27

3(x+4/3)^2 is always positive or at least 0 irrespective of the value of x
=> 3x² +8x-25 = 3(x + 4/3)^2 -27 cannot be smaller than -27.
The minimum is -27

When 3x² +8x-25 = 0
3(x + 4/3)^2 -27 = 0
3(x + 4/3)^2 = 27
(x+4/3)^2 = 9
x+4/3 = 3 or -3
x = 1 2/3 or -4 1/3

Answer 3

3x² +8x-25
= 3(x² + 8/3x) - 25
= 3(x² + 8/3x + 16/9 - 16/9) - 25
= 3[(x + 4/3)² - 16/9] - 25
= 3(x + 4/3)² - 16/3 - 25
= 3(x + 4/3)² - 91/3

a = 3, h = 4/3, k = -91/3

When x= -4/3, the minimum value is -91/3

3x² +8x-25=0
3(x + 4/3)² - 91/3 = 0
(x + 4/3)² = 91/9
x = -4/3 + Sqrt(91/9) or -4/3 - Sqrt(91/9)
x = (1/3)[-4 + sqrt(91)] or (1/3)[-4 - sqrt(91)]

Answer 4

I believe what you are asking is to simplify the problem.
3x^2+8x-25 can be broken down to:
x(3x+8)-25 so a=0 3x=-8 so x=-8/3

I hope thats what you are looking for. ill wait for response