Question

Solve by completing the square:
x^2+22x=10

Answer 1

what is half of 22?

Answer 2

1)Move the loose number over to the other side
2)Take half of the x-term (that is, divide it by two) (and don't forget the sign!), and square it. Add this square to both sides of the equation.
3)Convert the left-hand side to squared form. Simplify the right-hand side.
\[x ^{2}+22x=10\]
\[x^2+22x+(11)^2=10+(11)^2\]
\[(x+11)^2=131\]

Answer 3

so is the solution set the 11,131

Answer 4

ooops

Answer 5

Completing the square:
1. Write the quadratic equation as ax^2 + bx + c = 0
2. If the x^2 has a coefficient other than one, divide both sides of the equation by that coefficient.
3. Move the constant term (the number with no variable) to the right side.
4. Take half of the coefficient of x and square it.
5. Add the amount found in 3. to both sides.
6. The left side is now a perfect square, so write it a such.
7. \(X^2 = k\) has solutions \(X = \pm \sqrt k\)

Answer 6

8,-30

Answer 7

Let's follow each step with your problem.

The way your equation was given, you're already at the end of step 3.

1. Write the quadratic equation as ax^2 + bx + c = 0
2. If the x^2 has a coefficient other than one, divide both sides of the equation by that coefficient.
3. Move the constant term (the number with no variable) to the right side.

\(x^2+22x=10\)

4. Take half of the coefficient of x and square it.

The coefficient of x is 22. Half of that is 11. The square of 11 is 121.

5. Add the amount found in 3. to both sides.

\(x^2+22x + 121=10 + 121\)

\(x^2+22x + 121=131\)

6. The left side is now a perfect square, so write it a such.

\((x + 11)^2 = 131\)

7. \(X^2=k\) has solutions \(X=±\sqrt{k} \)

\(x + 11 = \pm \sqrt{131}\)

\(x + 11 = \sqrt{131}\) or \(x + 11 = -\sqrt{131}\)

\(x = -11+\sqrt{131}\) or \(x = -11-\sqrt{131}\)

Answer 8

BTW, in the instructions above, step 5 should read:

5. Add the amount found in 4. to both sides.

Answer 9

@hrassistant
Check the solution above. It's not 11,131 or 8,-30.