Solve the equation [tex]$x^2 - 16x + 54 = 0$[/tex] by completing the square.

Fill in the values of [tex]a[/tex] and [tex]b[/tex] to complete the solutions.

[tex]\[
\begin{array}{l}
x = a - \sqrt{b} \\
x = a + \sqrt{b}
\end{array}
\][/tex]



Answer :

To solve the quadratic equation [tex]\( x^2 - 16x + 54 = 0 \)[/tex] by completing the square, we will follow a detailed, step-by-step process. Let's break it down:

1. Start with the equation:
[tex]\[ x^2 - 16x + 54 = 0 \][/tex]

2. Rearrange the equation to isolate the constant term on one side:
[tex]\[ x^2 - 16x = -54 \][/tex]

3. Find the value to complete the square:
To complete the square, take the coefficient of [tex]\( x \)[/tex] (which is -16), divide it by 2, and then square it:
[tex]\[ \left(\frac{-16}{2}\right)^2 = (-8)^2 = 64 \][/tex]

4. Add and subtract this square within the equation:
[tex]\[ x^2 - 16x + 64 - 64 = -54 \][/tex]
Rearrange to make it easier:
[tex]\[ x^2 - 16x + 64 = 64 - 54 \][/tex]
[tex]\[ x^2 - 16x + 64 = 10 \][/tex]

5. Rewrite the left side as a perfect square trinomial:
[tex]\[ (x - 8)^2 = 10 \][/tex]

6. Take the square root of both sides:
[tex]\[ x - 8 = \pm \sqrt{10} \][/tex]

7. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 8 \pm \sqrt{10} \][/tex]

Given this method, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] can be identified from the solutions. The solutions in the form of [tex]\( x = a \pm \sqrt{b} \)[/tex] are:

[tex]\[ \begin{array}{l} x = 8 - \sqrt{10} \\ x = 8 + \sqrt{10} \end{array} \][/tex]

Thus, the values are:
[tex]\[ a = 8 \][/tex]
[tex]\[ b = 10 \][/tex]

So, to fill in the box:

[tex]\[ \begin{array}{l} x = 8 - \sqrt{10} \\ x = 8 + \sqrt{10} \end{array} \][/tex]

These are the solutions to the quadratic equation [tex]\( x^2 - 16x + 54 = 0 \)[/tex] by completing the square.