Question

How do you solve `y^2-90=13y`?

Answer 1

I used the quadratic formula, and found:

`y=18 or y=-5`

Explanation:

We have:

`y^2-90=13y`

Rearrange into standard form:

`y^2-13y-90=0`

Standard form has the powers of the pronumeral in decreasing order on the left and `0` on the right. It can be written as:

`ay^2+by+c=0`, where `a`, `b` and `c` are the coefficients of the powers of the pronumeral.

In this case, `a=1`, `b=-13` and `c=-90`.

We often use `x` as the pronumeral, but we treat `y` just the same way. The quadratic formula in terms of `y` is just:

`y=(-b+-sqrt(b^2-4ac))/(2a)`

Substituting in the values from the equation:

`y = (13+-sqrt((-13)^2-4(1)(-90)))/(2(1))`

`y = (13+-sqrt(169+360))/2 = (13+-23)/2`

Therefore:

`y=18 or y=-5`

This quadratic could also be solved by factorising.

Answer 2

`y=18 and -5`

Explanation:

Write as:`" "y^2-13y-90=0`

This is no different to a quadratic in `x` in format. The only difference is that it is in `y`. So the whole thing is:

`x=0=y^2-13y-90`

A quadratic in `x` where `ax^2>0` gives `uu`
A quadratic in `y` where `ay^2>0` gives `sub`
So it is rotating `uucolor(white)(.)90^o` clockwise.

Lets check the whole number factors of 90 to see if any of them give a difference of 13

`1xx90=90`
`2xx45=90`
`3xx30=90`
`4xx?=90larr" ? will not be a whole number factor"`
`color(purple)(5xx18=90 larr" difference is 13")`

The 90 is negative so the two numbers are of opposite sign.
The 13 is negative so the greater of the two is negative giving:

`(y-18)(y+5)=0`

Condition 1 : `" "y-18=0 " "=>" "y=+18`
Condition 2: `" "y+color(white)(1)5=0" "=>" "y=-color(white)(1)5`

Thus `y=18 and -5`