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

Step-by-step 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} \\ (2q^2)/(2)+(8q)/(2)-(3)/(2)=0 \\ q^2+4q-(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-(3)/(2)=0 \\ add\text{ 3/2 in both sides} \\ q^2+4q-(3)/(2)+(3)/(2)=+(3)/(2) \\ q^2+4q=(3)/(2) \end{gathered}`

Take half of the q term and square it

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

then add the result to both sides

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

Step 3

finally, isolate x

therefore, the solutions area

x=0.34

and

x=-4.34