Solving Quadratic Equations by the Square Root Method – College Algebra Study Guide

Q4. Solve: $2y^2 + 4 = 24$

Background

Topic: Quadratic Equations – Square Root Method

This question tests your ability to solve a quadratic equation by isolating the squared term and then applying the square root property.

Key Terms and Formulas

• Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$

• Square Root Property: If $x^2 = k$, then $x = \pm\sqrt{k}$

Step-by-Step Guidance

1. Subtract 4 from both sides to isolate the $y^2$ term: $2y^2 = 20$

2. Divide both sides by 2 to solve for $y^2$: $y^2 = 10$

3. Apply the square root property: $y = \pm\sqrt{10}$

Try solving on your own before revealing the answer!

Final Answer: $y = \pm\sqrt{10}$

We isolated $y^2$ and applied the square root property to get two solutions: $y = \sqrt{10}$ and $y = -\sqrt{10}$.

Q5. Solve: $(x + 3)^2 = 16$

Background

Topic: Quadratic Equations – Square Root Method

This question tests your ability to solve a quadratic equation by isolating the squared binomial and applying the square root property.

Key Terms and Formulas

• Square Root Property: If $x^2 = k$, then $x = \pm\sqrt{k}$

Step-by-Step Guidance

1. Take the square root of both sides: $x + 3 = \pm\sqrt{16}$

2. Recall that $\sqrt{16} = 4$, so $x + 3 = 4$ or $x + 3 = -4$

3. Solve each equation for $x$ by subtracting 3 from both sides.

Try solving on your own before revealing the answer!

Final Answer: $x = 1$ or $x = -7$

Solving $x + 3 = 4$ gives $x = 1$, and $x + 3 = -4$ gives $x = -7$.

Q6. Solve: $(4x - 5)^2 - 48 = 0$

Background

Topic: Quadratic Equations – Square Root Method

This question tests your ability to solve a quadratic equation by isolating the squared binomial and applying the square root property, including handling constants.

Key Terms and Formulas

• Square Root Property: If $x^2 = k$, then $x = \pm\sqrt{k}$

Step-by-Step Guidance

1. Add 48 to both sides to isolate the squared term: $(4x - 5)^2 = 48$

2. Take the square root of both sides: $4x - 5 = \pm\sqrt{48}$

3. Simplify $\sqrt{48}$ if possible, and then solve for $x$ by adding 5 and dividing by 4.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{5 \pm 4\sqrt{3}}{4}$

$\sqrt{48} = 4\sqrt{3}$, so $4x - 5 = \pm 4\sqrt{3}$. Solving for $x$ gives $x = \frac{5 + 4\sqrt{3}}{4}$ and $x = \frac{5 - 4\sqrt{3}}{4}$.