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Consider the equation:
[tex]\[ 0 = x^2 - 10x + 10 \][/tex]

1. Rewrite the equation by completing the square.

Your equation should look like [tex]\((x + a)^2 = b\)[/tex] or [tex]\((x - c)^2 = d\)[/tex].
[tex]\[ 0 = (x - 5)^2 - 15 \][/tex]
[tex]\[ (x - 5)^2 = 15 \][/tex]

2. What are the solutions to the equation?

Choose one answer:
(A) [tex]\( x = -5 \pm \sqrt{15} \)[/tex]
(B) [tex]\( x = 5 \pm \sqrt{15} \)[/tex]


Sagot :

Certainly! Let's go through the steps to solve the given equation [tex]\( x^2 - 10x + 10 = 0 \)[/tex] by completing the square and then finding its solutions.

### Step 1: Completing the Square

To complete the square, we follow these steps:

1. Start with the quadratic equation:
[tex]\[ x^2 - 10x + 10 = 0 \][/tex]

2. Move the constant term to the other side of the equation:
[tex]\[ x^2 - 10x = -10 \][/tex]

3. To complete the square, we take half of the coefficient of the [tex]\( x \)[/tex] term, square it, and add it to both sides. The coefficient of [tex]\( x \)[/tex] is [tex]\(-10\)[/tex], so half of this is [tex]\(-5\)[/tex], and squaring it gives us [tex]\(25\)[/tex]. Add and subtract [tex]\(25\)[/tex] on the left side:
[tex]\[ x^2 - 10x + 25 - 25 = -10 \][/tex]
Simplifies to:
[tex]\[ x^2 - 10x + 25 = 15 \][/tex]

4. Now the left side is a perfect square trinomial:
[tex]\[ (x - 5)^2 = 15 \][/tex]

Therefore, the equation in its completed square form is:
[tex]\[ (x - 5)^2 = 15 \][/tex]

### Step 2: Solving the Completed Square Equation

Next, we solve for [tex]\( x \)[/tex].

1. Take the square root of both sides:
[tex]\[ x - 5 = \pm \sqrt{15} \][/tex]

2. Solve for [tex]\( x \)[/tex] by isolating it:
[tex]\[ x = 5 \pm \sqrt{15} \][/tex]

Thus, the solutions to the equation are:
[tex]\[ x = 5 - \sqrt{15} \quad \text{and} \quad x = 5 + \sqrt{15} \][/tex]

### Conclusion

The solutions to the equation [tex]\( x^2 - 10x + 10 = 0 \)[/tex] are [tex]\(\boxed{x = 5 - \sqrt{15}} \)[/tex] and [tex]\(\boxed{x = 5 + \sqrt{15}} \)[/tex].

Therefore, the correct answer is:
(B) [tex]\( x = 5 \pm \sqrt{15} \)[/tex]