Question

directions: solve the following quadratic equations by square roots.

1. $x^2 - 16 = 0$
2. $x^2 - 100 = 0$
3. $x^2 + 25 = 0$
4. $x^2 + 7 = 88$
5. $x^2 - 5 = -4$
6. $6x^2 = 54$
7. $-2x^2 = -98$
8. $\frac{3}{4}x^2 = 48$
9. $3x^2 - 108 = 0$
10. $5x^2 - 45 = 0$

Explanation:

1. Equation 1: $x^2 -16=0$

Step1: Isolate $x^2$

$x^2 = 16$

Step2: Take square roots

$x = \pm\sqrt{16} = \pm4$

2. Equation 2: $x^2 -100=0$

Step1: Isolate $x^2$

$x^2 = 100$

Step2: Take square roots

$x = \pm\sqrt{100} = \pm10$

3. Equation 3: $x^2 +25=0$

Step1: Isolate $x^2$

$x^2 = -25$

Step2: Evaluate square roots

No real solutions (square of real number can't be negative)

4. Equation 4: $x^2 +7=88$

Step1: Isolate $x^2$

$x^2 = 88 - 7 = 81$

Step2: Take square roots

$x = \pm\sqrt{81} = \pm9$

5. Equation 5: $x^2 -5=-4$

Step1: Isolate $x^2$

$x^2 = -4 + 5 = 1$

Step2: Take square roots

$x = \pm\sqrt{1} = \pm1$

6. Equation 6: $6x^2=54$

Step1: Isolate $x^2$

$x^2 = \frac{54}{6} = 9$

Step2: Take square roots

$x = \pm\sqrt{9} = \pm3$

7. Equation 7: $-2x^2=-98$

Step1: Isolate $x^2$

$x^2 = \frac{-98}{-2} = 49$

Step2: Take square roots

$x = \pm\sqrt{49} = \pm7$

8. Equation 8: $\frac{3}{4}x^2=48$

Step1: Isolate $x^2$

$x^2 = 48 \times \frac{4}{3} = 64$

Step2: Take square roots

$x = \pm\sqrt{64} = \pm8$

9. Equation 9: $3x^2 -108=0$

Step1: Isolate $x^2$

$3x^2 = 108 \implies x^2 = \frac{108}{3} = 36$

Step2: Take square roots

$x = \pm\sqrt{36} = \pm6$

10. Equation 10: $5x^2 -45=0$

Step1: Isolate $x^2$

$5x^2 = 45 \implies x^2 = \frac{45}{5} = 9$

Step2: Take square roots

$x = \pm\sqrt{9} = \pm3$

Answer:

1. $x = 4$ or $x = -4$
2. $x = 10$ or $x = -10$
3. No real solutions
4. $x = 9$ or $x = -9$
5. $x = 1$ or $x = -1$
6. $x = 3$ or $x = -3$
7. $x = 7$ or $x = -7$
8. $x = 8$ or $x = -8$
9. $x = 6$ or $x = -6$
10. $x = 3$ or $x = -3$