Solve {l}{9b^2-9=279}{text{A)}{2sqrt{5},-2sqrt{5}}}{text{B)}{29,-29}}{text{C)}{4sqrt{2},-4sqrt{2}}}{text{D)}{sqrt{29},-sqrt{29}}}

$$\left. \begin{array} { l } { 9 b ^ { 2 } - 9 = 279 } \\ { \text { A) } \{ 2 \sqrt { 5 } , - 2 \sqrt { 5 } \} } \\ { \text { B) } \{ 29 , - 29 \} } \\ { \text { C) } \{ 4 \sqrt { 2 } , - 4 \sqrt { 2 } \} } \\ { \text { D) } \{ \sqrt { 29 } , - \sqrt { 29 } \} } \end{array} \right.$$

Solve for b

$$b=4\sqrt{2}\approx 5.656854249$$

$$b=-4\sqrt{2}\approx -5.656854249$$

Steps by Finding Square Root

Add 9 to both sides.

$$9b^{2}=279+9$$

Add 279 and 9 to get 288.

$$9b^{2}=288$$

Divide both sides by 9.

$$b^{2}=\frac{288}{9}$$

Divide 288 by 9 to get 32.

$$b^{2}=32$$

Take the square root of both sides of the equation.

$$b=4\sqrt{2} b=-4\sqrt{2}$$

Steps Using the Quadratic Formula

Subtract 279 from both sides.

$$9b^{2}-9-279=0$$

Subtract 279 from -9 to get -288.

$$9b^{2}-288=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -288 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 9\left(-288\right)}}{2\times 9}$$

Square 0.

$$b=\frac{0±\sqrt{-4\times 9\left(-288\right)}}{2\times 9}$$

Multiply -4 times 9.

$$b=\frac{0±\sqrt{-36\left(-288\right)}}{2\times 9}$$

Multiply -36 times -288.

$$b=\frac{0±\sqrt{10368}}{2\times 9}$$

Take the square root of 10368.

$$b=\frac{0±72\sqrt{2}}{2\times 9}$$

Multiply 2 times 9.

$$b=\frac{0±72\sqrt{2}}{18}$$

Now solve the equation b=\frac{0±72\sqrt{2}}{18} when ± is plus.

$$b=4\sqrt{2}$$

Now solve the equation b=\frac{0±72\sqrt{2}}{18} when ± is minus.

$$b=-4\sqrt{2}$$

The equation is now solved.

$$b=4\sqrt{2} b=-4\sqrt{2}$$