205,642 views
43 votes
43 votes
Need to solve for x by completing the sq and round to nearest 10th

Need to solve for x by completing the sq and round to nearest 10th-example-1
User Jminkler
by
2.5k points

1 Answer

14 votes
14 votes

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}

Need to solve for x by completing the sq and round to nearest 10th-example-1
User Harshal Chaudhari
by
2.4k points