QuestionSolve the equation by completing the square: $x^{2}-8 x+26=5$ . Find the roots.

Studdy Solution

STEP 1

Assumptions1. We are given the equation $x^{}-8 x+26=5$ . We need to solve this equation by completing the square

STEP 2

First, we rearrange the equation to get all terms on one side of the equation. This will give us a quadratic equation in standard form.
$x^{2}-8 x+26-5=0$

STEP 3

implify the equation.
$x^{2}-8 x+21=0$

STEP 4

In a quadratic equation of the form $ax^{2}+bx+c=0$ , the term $(b/2a)^{2}$ completes the square. In our equation, $a=1$ and $b=-8$ , so we calculate $(-8/21)^{2}$ .
$\left(\frac{-8}{21}\right)^{2} =16$

STEP 5

We rewrite the quadratic equation by adding and subtracting $(b/2a)^{2}$ inside the equation.
$x^{2}-8 x+16+21-16=0$

STEP 6

Rearrange the equation to group the terms that form a perfect square.
$(x^{2}-8 x+16) + (21-16) =0$

STEP 7

implify the equation. The first group of terms is a perfect square, and the second group of terms simplifies to5.
$(x-4)^{2} +5 =0$

STEP 8

Now, we can solve for $x$ by first isolating the square term.
$(x-4)^{2} = -5$

STEP 9

To get rid of the square, we take the square root of both sides. Remember that the square root of a square is the absolute value, so we have two solutions $x-4 = \sqrt{-5}$ and $x-4 = -\sqrt{-5}$ .
$x-4 = \sqrt{-5}, x-4 = -\sqrt{-5}$

STEP 10

We know that the square root of a negative number is an imaginary number. So, $\sqrt{-5}$ is equal to $i\sqrt{5}$ .
$x-4 = i\sqrt{5}, x-4 = -i\sqrt{5}$

STEP 11

Finally, we solve for $x$ in each equation.
$x =4 + i\sqrt{5}, x =4 - i\sqrt{5}$ So, the solutions to the equation $x^{}-8 x+26=5$ are $x =4 + i\sqrt{5}$ and $x =4 - i\sqrt{5}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms