Question

Please help me solve this and you have to solve each equation by completing the square with steps

Answer 1

x=0.34

and

x=-4.34

Explanation

Step 1

given

$2q^2+8q=3$

a)

$\begin{gathered} 2q^2+8q=3 \\ subtract\text{ 3 in boht sides} \\ 2q^2+8q-3=3-3 \\ 2q^2+8q-3=0 \end{gathered}$

now,

$\begin{gathered} 2q^{2}+8q-3=0 \\ divide\text{ bothsides by 2} \\ \frac{2q^2}{2}+\frac{8q}{2}-\frac{3}{2}=0 \\ q^2+4q-\frac{3}{2}=0 \end{gathered}$

Step 2

now,move

the constant to the right side by adding it on both sides

$\begin{gathered} q^{2}+4q-\frac{3}{2}=0 \\ add\text{ 3/2 in both sides} \\ q^2+4q-\frac{3}{2}+\frac{3}{2}=+\frac{3}{2} \\ q^2+4q=\frac{3}{2} \end{gathered}$

Take half of the q term and square it

$\begin{gathered} 4q\Rightarrow x\text{ term} \\ (4*\frac{1}{2})^2=4 \end{gathered}$

then add the result to both sides

$\begin{gathered} q^{2}+4q=\frac{3}{2} \\ q^2+4q+4=\frac{3}{2}+4 \\ (x+2)^2=\frac{11}{2} \end{gathered}$

Step 3

finally, isolate x

$\begin{gathered} (x+2)^{2}=\frac{11}{2} \\ square\text{ root in both sides} \\ \sqrt{(x+2)^2}=\sqrt{\frac{11}{2}} \\ x+2=\pm\sqrt{\frac{11}{2}} \\ subtract\text{ 2 in both sides} \\ x+2-2=\operatorname{\pm}\sqrt{\frac{11}{2}}-2 \\ x_1=\sqrt{\frac{11}{2}}-2=0.34 \\ x_2=-\sqrt{\frac{11}{2}}-2=-4.34 \end{gathered}$

therefore, the solutions area

x=0.34

and

x=-4.34