Question

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?

Answer 1

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`

Answer 2

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.

Answer 3

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
`" "rarr`completing 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.
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Remember that you may not divide by a variable unless you are sure it is not equal to `0`.