Q36. - Algebra 1

Question

Solve each equation by completing the square. Round to the nearest tenth if necessary.

$-3x^2+4=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are $\mathbf{-}\mathbf{1}\mathbf{.2}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}}\mathbf{1}\mathbf{.2}$. 

Step 1. Complete the square.

Complete the square to solve the quadratic equations by isolating and bx-terms.

 $$\begin{gathered} -3x^2+4=0 \\ -3x^2+4-4=0-4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\hspace{0.17em}\hspace{0.17em}subtract\hspace{0.17em}\hspace{0.17em}}4\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}\hspace{0.17em}each\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}]\hspace{0.17em}} \\ -3x^2=-4 \\ 3x^2=4 \end{gathered}$$

Step 2. Take the square root of each side.

Now take the square root of each side.

 $\begin{array}{c}\sqrt{3x^2}=\sqrt{4}\\ \sqrt{3}x=\pm 2\end{array}$

Step 3. Write the solutions.

Now divide both sides by $\sqrt{3}$ and separate the solutions.

 $$\begin{gathered} \sqrt{3}x=\pm 2 \\ \frac{\sqrt{3}x}{\sqrt{3}}=\pm \frac{2}{\sqrt{3}} \\ x=\frac{2}{\sqrt{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-\frac{2}{\sqrt{3}} \end{gathered}$$ 

Therefore, the solutions are $\frac{2}{\sqrt{3}}$ and $-\frac{2}{\sqrt{3}}$ which is equivalent to respectively.