Question

What kind of solutions does `5x^2 + 8x + 7 = 0` have?

Answer 1

Quadratic has no real solutions

Explanation:

The key realization is that our quadratic is of the form

`ax^2+bx+c=0`, where the value `b^2-4ac` is the discriminant, which tells us

`I.b^2-4ac >0=>2` real roots

`II.b^2-4ac<0=>` no real roots

`III.b^2-4ac=0=>1` real root

In our quadratic, we know `a=5, b=8` and `c=7`. So let's plug these values in. We get

`8^2-4(5*7)`

`=>64-4(35)`

`64-140=color(blue)(-76)`

Notice, our discriminant is less than zero, so we are in scenario `II`. This means our quadratic has no real solutions.

Hope this helps!

Answer 2

The roots are all complex numbers.

Explanation:

The quadratic equation is `x=(-b+-sqrt(b^2-4ac))/(2a)`.

Here, `a=5`, `b=8`, and `c=7`.

Substituting, we get `x=(-8+-sqrt(8^2-4*5*7))/(2*5)`.

Simplify.

`x=(-8+-sqrt(64-140))/(10)`

Here, we can already tell that the answers will be complex numbers. If you want to simplify further...

`x=(-8+-sqrt(-76))/(10)`

`x=(-8+-4sqrt(-19))/(10)`

`x=(-4+-2sqrt(-19))/(5)`