Question

What are the roots, real or imaginary, of the quadratic equation `y=8x^2-14x+15`?

Answer 1

The roots are the values for `x`.

`x=(7+isqrt71)/8,` `(7-isqrt71)/8`

Explanation:

You can use the quadratic formula.

`y=8x^2-14x+15` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where `a=8`, `b=-14`, and `c=15`.

The roots are the x-intercepts, which are the values for `x` when `y=0`, therefore, substitute `0` for `y`.

`0=8x^2-14x+15`

Quadratic Formula

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

Insert the given values into the formula.

`x=(-(-14)+-sqrt((-14)^2-4*8*15))/(2*8)`

Simplify.

`x=(14+-sqrt(196-480))/(16)`

`x=(14+-sqrt(-284))/16`

Prime factorize the radicand.

`x=(14+-sqrt(2xx2xx71xxi))/16`

Simplify.

`x=(14+-2isqrt71)/16`

Factor out the common factor `2`.

`x=(7+-isqrt71)/8`

Solutions for `x`.

`x=(7+isqrt71)/8,` `(7-isqrt71)/8`