Question

use the discriminant to determine the number and types of solutions of the quadratic equation
$4x^2 + 12x = -9$
the equation has
one real solution

two complex but not real solutions

two real solutions

Explanation:

Step1: Rewrite equation in standard form

A quadratic equation is in the form \(ax^{2}+bx + c = 0\). Given \(4x^{2}+12x=-9\), add 9 to both sides: \(4x^{2}+12x + 9 = 0\). Here, \(a = 4\), \(b = 12\), \(c = 9\).

Step2: Calculate the discriminant

The discriminant of a quadratic \(ax^{2}+bx + c = 0\) is \(D=b^{2}-4ac\). Substitute \(a = 4\), \(b = 12\), \(c = 9\) into the formula:
\(D=(12)^{2}-4\times4\times9\)
\(D = 144-144\)
\(D = 0\).

Step3: Determine the number and type of solutions

• If \(D>0\), two distinct real solutions.
• If \(D = 0\), one real solution (a repeated root).
• If \(D<0\), two complex (non - real) solutions.

Since \(D = 0\), the equation has one real solution.

Answer:

one real solution