Question

Using the discriminant, determine the number of real solutions.

Answer 1

Answer:

A

Step-by-step 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:

$\Delta=1-224=-223$

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

Hence, our answer is A.