Question

study the solutions of the three equations on the right. then, complete the statements below. there are two real solutions if the radicand is there is one real solution if the radicand is there are no real solutions if the radicand is done 1. y = -16x² + 32x - 10 x = (-32 ± √384)/(-32) 2. y = 4x² + 12x + 9 x = (-12 ± √0)/8 3. y = 3x² - 5x + 4 x = (5 ± √-23)/6

Explanation:

Step1: Analyze radicand for two real solutions

For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant (radicand in quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)) is \(D = b^2 - 4ac\). If \(D>0\), there are two real solutions. In the first equation, radicand is \(384>0\), so two real solutions when radicand is positive.

Step2: Analyze radicand for one real solution

If \(D = 0\), there is one real solution (a repeated root). In the second equation, radicand is \(0\), so one real solution when radicand is zero.

Step3: Analyze radicand for no real solutions

If \(D<0\), there are no real solutions (since square root of negative number is not real). In the third equation, radicand is \(- 23<0\), so no real solutions when radicand is negative.

Answer:

There are two real solutions if the radicand is positive.
There is one real solution if the radicand is zero.
There are no real solutions if the radicand is negative.