Question

Hi guys, umm so I wanted to know the solution to `x^2-15x+54=0`? I have the instructions to solve this but I don't understand it and I'm not good at anything math related

Answer 1

Given: `x^2-15x+54=0`

You can solve this equation by factoring.

Start by looking at the `x^2` term; this tells us that the first term in both factors is `x`:

`(xcolor(white)(-6))(xcolor(white)(-9))= 0`

Next, look that the `+54` term; it is positive, therefore, the signs are either both positive or both negative:

`(x+color(white)(6))(x+color(white)(9))= 0`

`(x-color(white)(6))(x-color(white)(9))= 0`

We can figure out which to choose by looking at `-15x`; it is negative, therefore, we should choose the case where the signs are both negative:

`(x-color(white)(6))(x-color(white)(9))= 0`

Note: If the middle term were positive then we would have chosen the "both signs are positive" case.

We need to figure out what numbers replace `r_1` and `r_2`:

`(x-r_1)(x-r_2)= 0`

We know that `r_1 xx r_2 = 54` and `r_1+r_2=15` the numbers `6` and `9` will satisfy that condition:

`(x-6)(x-9)= 0`

The product is 0 when either factor is 0, therefore, we set both factors equal to 0:

`x-6=0 and x-9= 0`

Solve both equations for x:

`x=6 and x=9`