Question

How do you solve x^2-11x+18=0

Answer 1

Answer: x = 9, 2

Explanation:

I'm going to give you two main ways to solve this type of problem.

First is doing it mentally:

When we're given the setting of ax^2 + bx + c

We want to find two numbers that multiply to c and add to b. (If a isn't 1, then we need to add that into our calculation, but in this case a = 1)

So, what two numbers multiply to 18 and add to -11?

-9 and -2, so we rewrite the equation as (x-9)(x-2)

We then solve for x, x - 9 = 0, so x = 9 and x - 2 = 0, so x = 2

Second is we use the quadratic formula:

It would be
`(-b+\sqrt{b^(2) -4ac})/2a\\(-b-\sqrt{b^(2) -4ac})/2a\\\\Plugging in your numbers: \\\\(11+\sqrt{11^(2) -72})/2\\(11-\sqrt{11^(2) -72})/2`

Answer 2

There are three ways of doing this. I will use the factorising method.
You come up with two numbers that times to get 18 and add together to get -11. The two numbers are -9 and -2. Then input them in the brackets. (x-2)(x-9)=0 and then because either one of them has to equal 0, you would have either x-2=0 or x-9=0 and you rearrange these equations to get either x=2 or x=9