Question

How do you solve `7k ^ { 2} - 48k - 64= 0`?

Answer 1

`k=8" "` or `" "k = -8/7`

Explanation:

Given:

`7k^2-48k-64 = 0`

We can solve this by completing the square then using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(7k-24)` and `B=32`

First multiply by `7` to make the leading term into a perfect square...

`0 = 7(7k^2-48k-64)`

`color(white)(0) = (7k)^2-2(7k)(24)-448`

`color(white)(0) = (7k)^2-2(7k)(24)+24^2-1024`

`color(white)(0) = (7k-24)^2-32^2`

`color(white)(0) = ((7k-24)-32)((7k-24)+32)`

`color(white)(0) = (7k-56)(7k+8)`

`color(white)(0) = 7(k-8)(7k+8)`

Hence `k=8` or `k = -8/7`

Answer 2

`-8/7` and 8

Explanation:

I use the new Transforming Method (Socratic, Google Search)
`f(k) = 7k^2 - 48k - 64 = 0.`
Proceeding: Solve the transformed equation:
`f'(k) = k^2 - 48k - 448 = 0`
to get its 2 real roots, then, divide them by a = 7.
Compose factor pairs of (ac = -448) --> ...(-4, 112)(-8, 56). This sum is (48 = - b). There fore, the 2 roots of f'(k) are: - 8 and 56.
Back to f(k), the 2 real roots are: `k1 = - 8/a = - 8/7`, and
`k2 = 56/a = 56/7 = 8.`
Answers: - 8/7, and 8.

NOTE . This method can avoid the lengthy factoring by grouping and solving the 2 binomials.