Question

How do you solve `-16x ^ { 2} = 12x ^ { 2} + 24x + 5`?

Answer 1

`x = -5/14, -1/2`

Explanation:

In order to solve this equation, first we will have to convert it into the general form, `ax^2 + bx + c = 0`, where `a != 0`.

So, `-16x^2 = 12x^2 + 24x + 5`

`rArr 28x^2 + 24x + 5 = 0` [Transposing `-16x^2` to the R.H.S]

Now, we will use the Quadratic Formula or Sridhar Acharya's Formula, to solve for the two roots.

Here, Discriminant = `D` = `b^2 - 4ac = (24)^2 - 4 * 28 * 5`

`= 576 - 560 = 16 gt 0`

So, the equation will have two real and distinct roots.

Now, Applying the Formula,

`alpha` = `(- b + sqrt(D))/(2a) = (- 24 + sqrt(16))/(2 * 28) = -20/56 = -5/14`

and, `beta` = `(-b - sqrt(D))/(2a) = (- 24 - 4)/(2 * 28) = -28/56 = -1/2`

So, the two roots of the equation are `-5/14` and `-1/2`.

Answer 2

See a solution process below:

Explanation:

First, add `color(red)(16x^2)` to each side of the equation to put the equation in standard form:

`-16x^2 + color(red)(16x^2) = 12x^2 + color(red)(16x^2) + 24x + 5`

`0 = (12 + color(red)(16))x^2 + 24x + 5`

`0 = 28x^2 + 24x + 5`

`28x^2 + 24x + 5 = 0`

Next. we can factor the left side of the equation as:

`(14x + 5)(2x + 1) = 0`

Now, we can solve each term on the left for `0`:

Solution 1:

`14x + 5 = 0`

`14x + 5 - color(red)(5) = 0 - color(red)(5)`

`14x + 0 = -5`

`14x = -5`

`(14x)/color(red)(14) = -5/color(red)(14)`

`(color(red)(cancel(color(black)(14)))x)/cancel(color(red)(14)) = -5/14`

`x = -5/14`

Solution 2:

`2x + 1 = 0`

`2x + 1 - color(red)(1) = 0 - color(red)(1)`

`2x + 0 = -1`

`2x = -1`

`(2x)/color(red)(2) = -1/color(red)(2)`

`(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -1/2`

`x = -1/2`

The Solution Is:

`x = {-1/2, -5/14}`