Question

Need to solve for x by completing the sq and round to nearest 10th

Answer 1

Question:

Solution:

Consider the following expression:

$x^2+20x-18=3$

this is equivalent to:

$x^2+20x=3+18$

this is equivalent to:

$x^2+20x=21$

Completing the square we get:

$x^2+20x+(\frac{b}{2a})^2=21+(\frac{b}{2a})^2$

here

a= 1

b = 20

then, we obtain:

$x^2+20x+(\frac{20}{2})^2=21+(\frac{20}{2})^2$

this is equivalent to:

$x^2+20x+(10)^2=21+(10)^2$

that is:

$x^2+20x+(10)^2=121$

notice that the left side of the equation is a perfect square, and therefore the equation becomes:

$(x+10)^2=121$

Now, to solve for x, we apply the square root to both sides of the equation and we get:

$x+10^{}=\pm\sqrt[]{121}$

solving for x, we get:

$x^{}=\pm\sqrt[]{121}-10=\pm11-10$

then, the correct answers are:

$x\text{ = 11-10 = 1}$

and

$x\text{ = -11 -10= -21}$

So that, the correct answer is:

$x\text{ = 1}$

and

$x\text{ = -21}$