Question

How to solve the equation by completing the square. (not sure what to do with the 2x at the end of the problem)

Answer 1

Answer:

x = 8, or -11.

Step-by-step explanation:

Given:

the given expression.

$2x^2+6x-14=0$

We need to solve the given expression by completing the square.

Solution:

Rewrite the expression as:

$2x^2+6x-14=0$

Whole expression divided by 2.

$x^2+3x-7=0$

Where:

$b=3\ and\ c=-7$

Solve the given expression by the following formula.

$x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c=0$ --------------(1)

Substitute b and c value in equation 1.

$(x+\frac{3}{2})^2-(\frac{3}{2})+(-7)=0$

$(x+\frac{3}{2})^2-\frac{3}{2}-7=0$

$(x+\frac{3}{2})^2+\frac{-3-7\times 2}{2}=0$

$(x+\frac{3}{2})^2+\frac{-3-14}{2}=0$

$(x+\frac{3}{2})^2+\frac{-19}{2}=0$

$(x+\frac{3}{2})^2-\frac{19}{2}=0$

Add $\frac{19}{2}$ both side of the equation.

$(x+\frac{3}{2})^2-\frac{19}{2}+\frac{19}{2}=\frac{19}{2}$

$(x+\frac{3}{2})^2=\frac{19}{2}$

Square root of the whole equation.

$\sqrt{(x+\frac{3}{2})^2}=\sqrt{\frac{19}{2}}$

$x+\frac{3}{2}=\pm\frac{19}{2}$ --------(1)

For positive sign.

Add $\frac{3}{2}$ both side of the equation 1.

$x+\frac{3}{2}-\frac{3}{2}=\frac{19}{2}-\frac{3}{2}$

$x=\frac{19}{2}-\frac{3}{2}$

$x=\frac{19-3}{2}$

$x=\frac{16}{2}$

x = 8

Similarly for negative sign.

x = -11

Therefore, the value of x = 8, or -11.