How do I solve this equation? y^2+35=8y

Do I divide the 35 by y as well as the y^2 and the 8y? Or maybe I solve it as a quadratic?

3 Answers
Mar 26, 2017

See below.

Explanation:

Mind that

(y-a)^2+b^2=y^2-2ay+a^2+b^2

now comparing with

y^2-8y+35

we have

-2a=-8->a=4

a^2+b^2=35->16+b^2=35->b^2=-29

At this point we have a real impossibility so the equation has no solution for real values.

Trying the complex domain we can state b = pm i sqrt29.

In the complex domain, the solution arise from

(y+4)^2+29=0 or

y+4 = pm i sqrt 29 and finally

y = -4pm isqrt 29

Mar 26, 2017

You do not solve this equation by division. This is a quadratic equation ; there are several ways to solve it. I will explain a solution below.

Explanation:

Given: y^2+35=8y

Subtract 8y from both sides of the equation:

y^2-8y+35=0

This is in a special case of the standard form

x=ay^2-by+c

where x = 0, a = 1, b = -8 and c = 35

There is something called a discriminant that will tell you whether this equation has, 0, 1, or 2 solutions.

The discriminant is:

d = b^2-4(a)(c)

Substituting in our special values:

d = (-8)^2-4(1)(35)

d = 64-140

d = -76

Because the discriminant is negative, we know that the equation has no real solutions; its solutions are a complex conjugate pair. I suspect that you have not, yet, learned about the complex number system so we must declare that this quadratic has no solutions.

Mar 26, 2017

This equation has no solution.

Explanation:

As soon as you have y^2 in an equation, you have a quadratic and you make it equal to 0:

y^2 -8y +35 =0

There are now 3 options:
" "rarr factorise
" "rarrcompleting the square
" "rarr use the quadratic formula

As this does not factorise, I will choose to complete the square.

y^2 -8y +" ? " = -35

y^2 -8y +color(red)((8/2)^2) = -35 +color(red)((8/2)^2)

(y-4)^2= -35+16

y-4 = +-sqrt(-19)

And here we have a problem, because sqrt(-19) is an imaginary number.

This equation has no solution.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Remember that you may not divide by a variable unless you are sure it is not equal to 0.