Basic Square Root Solution

Solve: $x^2 = 25$

Identify the equation form: The equation is already in the form $x^2 = k$ where $k = 25$ = Ready to take square root

Take the square root of both sides: $\sqrt{x^2} = \pm\sqrt{25}$ = $x = \pm\sqrt{25}$

Simplify the square root: $\sqrt{25} = 5$ = $x = \pm 5$

Write both solutions: $x = 5$ or $x = -5$ = Two solutions

Answer: $x = 5$ or $x = -5$

Isolate the Squared Term First

Solve: $3x^2 = 48$

Isolate $x^2$: Divide both sides by 3: $\frac{3x^2}{3} = \frac{48}{3}$ = $x^2 = 16$

Take the square root of both sides: $\sqrt{x^2} = \pm\sqrt{16}$ = $x = \pm\sqrt{16}$

Simplify: $\sqrt{16} = 4$ = $x = \pm 4$

Verify: Check: $3(4)^2 = 3 \times 16 = 48$ ✓ and $3(-4)^2 = 3 \times 16 = 48$ ✓ = Both solutions work

Answer: $x = 4$ or $x = -4$

Equation with Addition

Solve: $x^2 + 5 = 41$

Isolate $x^2$: Subtract 5 from both sides: $x^2 + 5 - 5 = 41 - 5$ = $x^2 = 36$

Take the square root of both sides: $\sqrt{x^2} = \pm\sqrt{36}$ = $x = \pm\sqrt{36}$

Simplify: $\sqrt{36} = 6$ = $x = \pm 6$

State both solutions: $x = 6$ or $x = -6$ = Two solutions

Answer: $x = 6$ or $x = -6$

Non-Perfect Square Result

Solve: $x^2 = 20$

Take the square root of both sides: $\sqrt{x^2} = \pm\sqrt{20}$ = $x = \pm\sqrt{20}$

Simplify the radical: $\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$ = $x = \pm 2\sqrt{5}$

Approximate if needed: $2\sqrt{5} \approx 2 \times 2.236 \approx 4.47$ = $x \approx \pm 4.47$

Answer: $x = \pm 2\sqrt{5}$ (approximately $\pm 4.47$)

Squared Binomial

Solve: $(x - 3)^2 = 49$

Take the square root of both sides: $\sqrt{(x-3)^2} = \pm\sqrt{49}$ = $x - 3 = \pm 7$

Solve for positive root: $x - 3 = 7 \implies x = 7 + 3$ = $x = 10$

Solve for negative root: $x - 3 = -7 \implies x = -7 + 3$ = $x = -4$

Verify: $(10-3)^2 = 7^2 = 49$ ✓ and $(-4-3)^2 = (-7)^2 = 49$ ✓ = Both solutions verified

Answer: $x = 10$ or $x = -4$

No Real Solution

Solve: $x^2 = -9$

Analyze the equation: We need a number whose square is $-9$ = Consider if this is possible

Recall the property of squares: Any real number squared is non-negative: $a^2 \geq 0$ for all real $a$ = Squares are always $\geq 0$

Conclude: No real number squared gives a negative result = No real solution

Answer: No real solution (the square of any real number cannot be negative)

Mistake: Forgetting the negative solution

Why: Students often write only the positive square root, forgetting that $(-a)^2 = a^2$ as well.

Correct: Always write $x = \pm\sqrt{k}$ to indicate both solutions. For $x^2 = 9$, write $x = \pm 3$, not just $x = 3$.

Mistake: Taking the square root before isolating $x^2$

Why: In equations like $2x^2 = 18$, students sometimes try to take $\sqrt{2x^2}$ directly.

Correct: First divide by 2 to get $x^2 = 9$, then take the square root: $x = \pm 3$.

Mistake: Thinking $\sqrt{x^2} = x$

Why: This is only true for positive $x$. For negative $x$, $\sqrt{x^2} = |x|$.

Correct: Write $\sqrt{x^2} = \pm x$ or more precisely $|x|$, which gives both positive and negative possibilities.

Mistake: Attempting to solve $x^2 = -k$ (negative) as a real number

Why: Students may try to write $x = \pm\sqrt{-k}$, which is not a real number.

Correct: Recognize that if $k < 0$, the equation $x^2 = k$ has no real solution.

Calculating Distances

In physics, the relationship between distance, acceleration, and time involves squared terms.

A ball is dropped and falls according to $d = 5t^2$ (in meters). How long does it take to fall 80 meters? Solve $5t^2 = 80$, so $t^2 = 16$, giving $t = 4$ seconds (we use only the positive value since time cannot be negative).

Finding Side Lengths from Area

When you know the area of a square, you can find the side length using square roots.

A square garden has an area of 144 square meters. The side length is $s$ where $s^2 = 144$, so $s = 12$ meters.

Projectile Motion

The height of a thrown ball involves quadratic equations.

A ball thrown upward reaches height $h = 64 - 16t^2$ feet. When is the ball at 48 feet? Solve $64 - 16t^2 = 48$, giving $16t^2 = 16$, so $t^2 = 1$ and $t = 1$ second.

Q: Why do we write $\pm$ (plus-minus)?

A: Because both a positive and negative number, when squared, give the same positive result. For example, $5^2 = 25$ and $(-5)^2 = 25$. So if $x^2 = 25$, both $x = 5$ and $x = -5$ are valid solutions.

Q: When does this method NOT work?

A: This method works best when the equation can be written as $x^2 = k$ or $(\text{expression})^2 = k$. If the equation has an $x$ term (like $x^2 + 3x = 10$), you'll need factoring or the quadratic formula instead.

Q: What if the answer is not a perfect square?

A: You can leave the answer in radical form (like $x = \pm\sqrt{7}$) or approximate with a calculator. Both forms are correct; the radical form is exact.