Question

use the discriminant to answer the questions.
32x - 4 = 4x² + 60
for the equation shown, choose the description of
the solutions.
○ one real, rational
○ two real, rational
○ two real, irrational
○ two complex

Explanation:

Step1: Rewrite in standard form

First, rewrite the quadratic equation \(32x - 4 = 4x^2 + 60\) in standard form \(ax^2 + bx + c = 0\). Subtract \(32x\) and add \(4\) to both sides: \(4x^2 - 32x + 64 = 0\). We can simplify by dividing by \(4\): \(x^2 - 8x + 16 = 0\). Here, \(a = 1\), \(b = -8\), \(c = 16\).

Step2: Calculate the discriminant

The discriminant of a quadratic \(ax^2 + bx + c\) is \(D = b^2 - 4ac\). Substitute \(a = 1\), \(b = -8\), \(c = 16\): \(D = (-8)^2 - 4(1)(16) = 64 - 64 = 0\). Wait, no—wait, original equation before simplifying: wait, let's check again. Wait, original equation: \(32x - 4 = 4x^2 + 60\). Moving all terms to right: \(0 = 4x^2 - 32x + 64\), which is \(4x^2 - 32x + 64 = 0\). Divide by \(4\): \(x^2 - 8x + 16 = 0\). Then discriminant \(D = (-8)^2 - 4(1)(16) = 64 - 64 = 0\)? Wait, no, that would be one real solution. But wait, maybe I made a mistake. Wait, let's re-express the original equation correctly. Wait, \(32x - 4 = 4x^2 + 60\) → \(4x^2 - 32x + 64 = 0\) (subtract \(32x\) and add \(4\) to both sides: \(4x^2 - 32x + 64 = 0\)). Then \(a = 4\), \(b = -32\), \(c = 64\). Then discriminant \(D = (-32)^2 - 4(4)(64) = 1024 - 1024 = 0\). Wait, but if discriminant is zero, there is one real rational solution? But wait, the options include "One real, rational". Wait, but let's check the simplification. Wait, \(4x^2 - 32x + 64 = 0\) can be factored as \(4(x^2 - 8x + 16) = 0\) → \(4(x - 4)^2 = 0\), so \(x = 4\) (a repeated root, so one real rational solution). But wait, the options have "One real, rational" as an option. Wait, but let's check again. Wait, original equation: \(32x - 4 = 4x^2 + 60\). Let's move all terms to left: \(-4x^2 + 32x - 64 = 0\), multiply by -1: \(4x^2 - 32x + 64 = 0\), same as before. So discriminant \(D = b^2 - 4ac = (-32)^2 - 4464 = 1024 - 1024 = 0\). So discriminant is zero, which means one real rational solution (since the root is a perfect square, rational). Wait, but the options: "One real, rational" is an option. Wait, but maybe I made a mistake in the equation. Wait, let's check the original problem again. The equation is \(32x - 4 = 4x^2 + 60\). Let's rearrange: \(4x^2 - 32x + 64 = 0\). Divide by 4: \(x^2 - 8x + 16 = 0\), which factors to \((x - 4)^2 = 0\), so \(x = 4\) (a single real rational solution). So the correct option is "One real, rational". Wait, but let's confirm the discriminant rules:

• If \(D > 0\) and a perfect square: two real rational.
• If \(D > 0\) and not a perfect square: two real irrational.
• If \(D = 0\): one real rational (repeated root).
• If \(D < 0\): two complex.

In this case, \(D = 0\), so one real rational solution.

Answer:

One real, rational (the option with "One real, rational")