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

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Answer 1 · 2018-06-25T22:09:12

$x=3+-isqrt2$

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

$x^2-6x=-11$

When we complete the square of an equation of the form

$ax^2+bx=-c$

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

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

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

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

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

Taking the square root of both sides gives us

$x-3=sqrt(-2)$

Which can be rewritten as

$x-3=sqrt(-1)*sqrt2$

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

$x-3=+-isqrt2$

Adding $3$ to both sides gives us

$x=3+-isqrt2$

Hope this helps!

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