Question

For the following quadratic functions, find the value(s) of x for the given value of y:
a) y=xˆ2+6x+10 when y=1
could you explain how to do this? thanks

Answer 1

So we have 1=x^2+6x+10

Answer 2

Now to solve quadratic equations you must have one side equal to 0.

Answer 3

Do you know how to get one of the sides of the equal sign to be zero?

Answer 4

well could we minus the 1 from the left side of the equation?

Answer 5

Yep. Remember whatever you do to one side, you must do to the other.

Answer 6

So what is the resulting equation after subtracting one from both sides?

Answer 7

0=xˆ2+6x+9

Answer 8

Perfect.
Do you know how to factor?

Answer 9

no

Answer 10

What about using the quadratic formula?

Answer 11

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 12

wait with factor is it factorizing?
would it give me 0=(x+3)(x+3) ?
if not then i don't know about using the quadratic formula either.

Answer 13

You factored correctly! :)
Now set each of those factors =0
like so
x+3=0 or x+3=0

This is the same equation
They will have the same solution

Answer 14

oh so it gives us -3?

Answer 15

Yep :)

Answer 16

wow okay thank you! (: