Question

Using the discriminant, determine the number of real solutions.

-2x2+x-28 = 0
O A. no real solutions
O B. one real solution
O C. two real solutions

Answer 1

A

Explanation:

For a quadratic equation in standard form:

`ax^2+bx+c=0`

The discriminant, symoblized as Δ, is given by:

`\Delta=b^2-4ac`

If:

• Δ>0 (positive), then our quadratic has two real roots.
• Δ<0 (negative), then our quadratic has no real roots. It does, however, have two imaginary (complex) roots.
• And if Δ=0 (zero), then our quadratic has exactly one real root.

We have the equation:

`-2x^2+x-28=0`

Thus:

`a=-2, b=1, c=-28`

Substituting them for our discriminant:

`\Delta=(1)^2-(4)(-2)(-28)`

Evaluate:

`\Delta=1-224=-223<0`

Since our discriminant is negative, there is no real solutions.

Hence, our answer is A.