How do you solve $x^{2} - 6 x + 11 = 0$ $x^2-6x+11=0$ by completing the square?

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Answer 1 · 2018-06-25T22:09:12

$x = 3 \pm i \sqrt{2}$ $x=3+-isqrt2$

Let's first subtract $11$ $11$ from both sides to get

$x^{2} - 6 x = - 11$ $x^2-6x=-11$

When we complete the square of an equation of the form

$a x^{2} + b x = - c$ $ax^2+bx=-c$

We take half of our $b$ $b$ value, square it, and add it to both sides.

Our $b$ $b$ value is $- 6$ $-6$ , half of that is $- 3$ $-3$ , and that value squared is $9$ $9$ . Adding it to the left and right gives us

$x^{2} - 6 x + 9 = - 11 + 9$ $color(steelblue)(x^2-6x+9)=-11+9$

What I have in blue can be factored as ${(x - 3)}^{2}$ $(x-3)^2$ . This gives us

${(x - 3)}^{2} = - 2$ $(x-3)^2=-2$

Taking the square root of both sides gives us

$x - 3 = \sqrt{- 2}$ $x-3=sqrt(-2)$

Which can be rewritten as

$x - 3 = \sqrt{- 1} \cdot \sqrt{2}$ $x-3=sqrt(-1)*sqrt2$

Note that $i = \sqrt{- 1}$ $i=sqrt(-1)$ . With this definition in mind, we now have

$x - 3 = \pm i \sqrt{2}$ $x-3=+-isqrt2$

Adding $3$ $3$ to both sides gives us

$x = 3 \pm i \sqrt{2}$ $x=3+-isqrt2$

Hope this helps!

How do you solve x2−6x+11=0x^2-6x+11=0 by completing the square?