Question

How do you use the quadratic formula to solve `4x^2+16x+15=0`?

Answer 1

See explanation...

Explanation:

`4x^2+16x+15 = 0`

is in the form:

`ax^2+bx+c = 0`

with `a=4`, `b=16` and `c=15`

The roots are given by the quadratic formula:

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

`color(white)(x) = (-16+-sqrt(16^2-4(4)(15)))/(2*4)`

`color(white)(x) = (-16+-sqrt(256-240))/8`

`color(white)(x) = (-16+-sqrt(16))/8`

`color(white)(x) = (-16+-4)/8`

`color(white)(x) = (-4+-1)/2`

That is:

`x = -5/2` or `x = -3/2`

`color(white)()`
Footnote

In this particular example, I would probably have favoured an AC method rather than the quadratic formula.

Given:

`4x^2+16x+15 = 0`

Find a pair of factors of `AC=4*15=60` with sum `B=16`

The pair `10, 6` works.

Use this pair to split the middle term and factor by grouping:

`0 = 4x^2+16x+15`

`color(white)(0) = (4x^2+10x)+(6x+15)`

`color(white)(0) = 2x(2x+5)+3(2x+5)`

`color(white)(0) = (2x+3)(2x+5)`

Hence roots `x = -3/2` and `x = -5/2`

Note that the AC method is only applicable if the quadratic has rational zeros, whereas the quadratic formula will always give you the solution, even if the roots are Complex.