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+((b)/(2a))^2=21+((b)/(2a))^2`

here

a= 1

b = 20

then, we obtain:

`x^2+20x+((20)/(2))^2=21+((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}`