Solving Quadratic Equations by Square Root Property │Algebra - Free Printable

To solve the given quadratic equations using the square root property, we will follow these steps:

1. Isolate the squared term on one side of the equation.
2. Take the square root of both sides, remembering to include both the positive and negative roots.
3. Solve for the variable.

Let's solve each equation step by step.

---

a) \( x^2 = 49 \)

1. Take the square root of both sides:
\[
x = \pm \sqrt{49}
\]
2. Simplify:
\[
x = \pm 7
\]

Solution:
\[
x = 7 \quad \text{or} \quad x = -7
\]

---

b) \( 4x^2 = 100 \)

1. Divide both sides by 4 to isolate \( x^2 \):
\[
x^2 = \frac{100}{4} = 25
\]
2. Take the square root of both sides:
\[
x = \pm \sqrt{25}
\]
3. Simplify:
\[
x = \pm 5
\]

Solution:
\[
x = 5 \quad \text{or} \quad x = -5
\]

---

c) \( x^2 - 72 = 0 \)

1. Add 72 to both sides to isolate \( x^2 \):
\[
x^2 = 72
\]
2. Take the square root of both sides:
\[
x = \pm \sqrt{72}
\]
3. Simplify \( \sqrt{72} \):
\[
\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}
\]
4. Therefore:
\[
x = \pm 6\sqrt{2}
\]

Solution:
\[
x = 6\sqrt{2} \quad \text{or} \quad x = -6\sqrt{2}
\]

---

d) \( 2x^2 + 14 = 39 \)

1. Subtract 14 from both sides:
\[
2x^2 = 25
\]
2. Divide both sides by 2 to isolate \( x^2 \):
\[
x^2 = \frac{25}{2}
\]
3. Take the square root of both sides:
\[
x = \pm \sqrt{\frac{25}{2}}
\]
4. Simplify:
\[
\sqrt{\frac{25}{2}} = \frac{\sqrt{25}}{\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}
\]
5. Therefore:
\[
x = \pm \frac{5\sqrt{2}}{2}
\]

Solution:
\[
x = \frac{5\sqrt{2}}{2} \quad \text{or} \quad x = -\frac{5\sqrt{2}}{2}
\]

---

e) \( \frac{5}{2}x^2 + 69 = 44 \)

1. Subtract 69 from both sides:
\[
\frac{5}{2}x^2 = -25
\]
2. Divide both sides by \( \frac{5}{2} \) (or multiply by \( \frac{2}{5} \)):
\[
x^2 = -25 \cdot \frac{2}{5} = -10
\]
3. Since \( x^2 = -10 \) has no real solutions (the square of a real number cannot be negative), the solution is:
\[
\text{No real solutions}
\]

Solution:
\[
\text{No real solutions}
\]

---

f) \( (x - 6)^2 = 81 \)

1. Take the square root of both sides:
\[
x - 6 = \pm \sqrt{81}
\]
2. Simplify:
\[
x - 6 = \pm 9
\]
3. Solve for \( x \):
\[
x = 6 + 9 \quad \text{or} \quad x = 6 - 9
\]
\[
x = 15 \quad \text{or} \quad x = -3
\]

Solution:
\[
x = 15 \quad \text{or} \quad x = -3
\]

---

g) \( (2x - 3)^2 = 121 \)

1. Take the square root of both sides:
\[
2x - 3 = \pm \sqrt{121}
\]
2. Simplify:
\[
2x - 3 = \pm 11
\]
3. Solve for \( x \):
\[
2x - 3 = 11 \quad \text{or} \quad 2x - 3 = -11
\]
\[
2x = 14 \quad \text{or} \quad 2x = -8
\]
\[
x = 7 \quad \text{or} \quad x = -4
\]

Solution:
\[
x = 7 \quad \text{or} \quad x = -4
\]

---

h) \( 3(x + 7)^2 + 11 = 59 \)

1. Subtract 11 from both sides:
\[
3(x + 7)^2 = 48
\]
2. Divide both sides by 3:
\[
(x + 7)^2 = 16
\]
3. Take the square root of both sides:
\[
x + 7 = \pm \sqrt{16}
\]
4. Simplify:
\[
x + 7 = \pm 4
\]
5. Solve for \( x \):
\[
x + 7 = 4 \quad \text{or} \quad x + 7 = -4
\]
\[
x = -3 \quad \text{or} \quad x = -11
\]

Solution:
\[
x = -3 \quad \text{or} \quad x = -11
\]

---

i) \( x^2 + 8x + 16 = 25 \)

1. Recognize that \( x^2 + 8x + 16 \) is a perfect square:
\[
(x + 4)^2 = 25
\]
2. Take the square root of both sides:
\[
x + 4 = \pm \sqrt{25}
\]
3. Simplify:
\[
x + 4 = \pm 5
\]
4. Solve for \( x \):
\[
x + 4 = 5 \quad \text{or} \quad x + 4 = -5
\]
\[
x = 1 \quad \text{or} \quad x = -9
\]

Solution:
\[
x = 1 \quad \text{or} \quad x = -9
\]

---

j) \( 9x^2 - 12x + 4 = 36 \)

1. Recognize that \( 9x^2 - 12x + 4 \) is a perfect square:
\[
(3x - 2)^2 = 36
\]
2. Take the square root of both sides:
\[
3x - 2 = \pm \sqrt{36}
\]
3. Simplify:
\[
3x - 2 = \pm 6
\]
4. Solve for \( x \):
\[
3x - 2 = 6 \quad \text{or} \quad 3x - 2 = -6
\]
\[
3x = 8 \quad \text{or} \quad 3x = -4
\]
\[
x = \frac{8}{3} \quad \text{or} \quad x = -\frac{4}{3}
\]

Solution:
\[
x = \frac{8}{3} \quad \text{or} \quad x = -\frac{4}{3}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{a) } x = 7 \text{ or } x = -7 \\
&\text{b) } x = 5 \text{ or } x = -5 \\
&\text{c) } x = 6\sqrt{2} \text{ or } x = -6\sqrt{2} \\
&\text{d) } x = \frac{5\sqrt{2}}{2} \text{ or } x = -\frac{5\sqrt{2}}{2} \\
&\text{e) No real solutions} \\
&\text{f) } x = 15 \text{ or } x = -3 \\
&\text{g) } x = 7 \text{ or } x = -4 \\
&\text{h) } x = -3 \text{ or } x = -11 \\
&\text{i) } x = 1 \text{ or } x = -9 \\
&\text{j) } x = \frac{8}{3} \text{ or } x = -\frac{4}{3}
\end{aligned}
}
\]