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Solve the quadratic equation by completing the square.x^2-10x+23=0First Tuesday appropriate form and fill in the blanks with the correct numbers. Then solve the equation. If there’s more than one solution separate them with commas.

Solve the quadratic equation by completing the square.x^2-10x+23=0First Tuesday appropriate-example-1
User Erubiel
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1 Answer

15 votes
15 votes

Answer:

  • Form: (x -5)² -2
  • Solution: x = 5+√2, 5-√2

Explanation:

You want to solve the quadratic x² -10x +23 = 0 by completing the square.

Form

The constant term in a perfect square trinomial is the square of half the coefficient of the linear term. Here, the linear term is -10x, so the constant will need to be (-10/2)² = 25. It is 23, so we can add and subtract 2 to get the form we need:

x² -10x +23 +2 -2 = 0

x² -10x +25 -2 = 0

(x -5)² -2 = 0 . . . . . . . . . . the equation in vertex form

Solution

We can add 2 and take the square root to find the solutions.

(x -5)² = 2 . . . . . . . . add 2 to both sides

x -5 = ±√2 . . . . . . . square root

x = 5 ± √2 . . . . . . add 5

The two solutions are ...

5 +√2, 5 -√2

__

Additional comments

If you don't want to think too hard about how to get the "k" in the vertex form (x -h)² +k, you can simply add and subtract the constant you found you need (25). This looks like ...

x² -10x +25 +23 -25 = 0

(x -5)² -2 = 0 . . . . as above

Sometimes this is easier than thinking, "I need 25 and I have 23, so I need to add 2 (and subtract 2)."

We add and subtract the same number so we don't change equation. Equivalently, we can add the same number on both sides of the equation:

x² -10x +23 +2 = 2 . . . . . 2 added to both sides

(x -5)² = 2 . . . . . . . . . . . a correct equation, but not the expression this problem is looking for in the "Form" section.

You will note the sign (minus) in the binomial square (x-5)² is the same as the sign of the x-coefficient (-10x).

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User Ranjith V
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3.4k points