How do you find the zeros, real and imaginary, of $y=(x-3)^2+41$ using the quadratic formula?

Question

How do you find the zeros, real and imaginary, of $y=(x-3)^2+41$ using the quadratic formula?

1 Answer

Impact of this question

1715 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-13T03:49:32

$x=3+sqrt(41)i,$ $3-sqrt(41)i$

Explanation:

Given:

$y=(x-3)^2+41$

Expand $(x-3)^2$ .

$y=x^2-6x+9+41$

Simplify.

$y=x^2-6x+50$ is a quadratic equation in standard form:

$y=ax^2+bx+c$ ,

where:

$a=1$ , $b=-6$ , and $c=50$

The zeroes are the x-intercepts, which are the values of $x$ when $y=0$ .

Substitute $0$ for $y$ .

$0=x^2-6x+50$

Quadratic Formula

$x=(-b+-sqrt((b)^2-4ac))/(2a)$

Plug in the known values.

$x=(-(-6)+-sqrt((-6)^6-4*1*50))/(2*1)$

Simplify.

$x=(6+-sqrt(-164))/2$

Prime factorize $164$ .

$x=(6+-sqrt(2xx2xx41))/2$

Simplify.

$x=(6+-2sqrt(41)i)/2$

Reduce.

$x=(color(red)cancel(color(black)(6))^3+-color(red)cancel(color(black)(2))^1sqrt(41)i)/color(red)cancel(color(black)(2))^1$

Simplify.

$x=3+-sqrt(41)i$

Values for $x$ .

$x=3+sqrt(41)i,$ $3-sqrt(41)i$

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of $y=(x-3)^2+41$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=(x-3)^2+41 y=(x-3)^2+41 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the zeros, real and imaginary, of $y=(x-3)^2+41$ using the quadratic formula?