Question

name: _____________period: ______algebraunit 8 hw 3 - finding the zeros with the square root methoddirections: show all work in order to receive full credit.1. solve: $-7n^2 = -448$2. solve: $n^2 - 5 = -4$3. solve: $4r^2 + 1 = 325$4. solve: $8b^2 - 4 = 532$5. solve: $9n^2 + 10 = 91$6. solve: $2m^2 + 10 = 210$

Explanation:

1. Solve $-7n^2=-448$

Step1: Isolate $n^2$

Divide both sides by $-7$:
$\frac{-7n^2}{-7}=\frac{-448}{-7}$
$n^2=64$

Step2: Take square roots

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

2. Solve $n^2 - 5=-4$

Step1: Isolate $n^2$

Add 5 to both sides:
$n^2 - 5 + 5=-4 + 5$
$n^2=1$

Step2: Take square roots

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

3. Solve $4r^2 + 1=325$

Step1: Isolate $4r^2$

Subtract 1 from both sides:
$4r^2 + 1 - 1=325 - 1$
$4r^2=324$

Step2: Isolate $r^2$

Divide both sides by 4:
$\frac{4r^2}{4}=\frac{324}{4}$
$r^2=81$

Step3: Take square roots

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

4. Solve $8b^2 - 4=532$

Step1: Isolate $8b^2$

Add 4 to both sides:
$8b^2 - 4 + 4=532 + 4$
$8b^2=536$

Step2: Isolate $b^2$

Divide both sides by 8:
$\frac{8b^2}{8}=\frac{536}{8}$
$b^2=67$

Step3: Take square roots

$b=\pm\sqrt{67}$

5. Solve $9n^2 + 10=91$

Step1: Isolate $9n^2$

Subtract 10 from both sides:
$9n^2 + 10 - 10=91 - 10$
$9n^2=81$

Step2: Isolate $n^2$

Divide both sides by 9:
$\frac{9n^2}{9}=\frac{81}{9}$
$n^2=9$

Step3: Take square roots

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

6. Solve $2m^2 + 10=210$

Step1: Isolate $2m^2$

Subtract 10 from both sides:
$2m^2 + 10 - 10=210 - 10$
$2m^2=200$

Step2: Isolate $m^2$

Divide both sides by 2:
$\frac{2m^2}{2}=\frac{200}{2}$
$m^2=100$

Step3: Take square roots

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

Answer:

1. $n = 8$ or $n = -8$
2. $n = 1$ or $n = -1$
3. $r = 9$ or $r = -9$
4. $b = \sqrt{67}$ or $b = -\sqrt{67}$
5. $n = 3$ or $n = -3$
6. $m = 10$ or $m = -10$