SOLUTION: Here I go again.....now, with negatives in the problem...I'm stumped!!! I am trying to help my daughter find the root of a quadratic equation using standard form. y=x^2-17x+30. Tha

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Question 432892: Here I go again.....now, with negatives in the problem...I'm stumped!!! I am trying to help my daughter find the root of a quadratic equation using standard form. y=x^2-17x+30. Thanks for your input!!!
Found 2 solutions by sudhanshu_kmr, Edwin McCravy:
Answer by sudhanshu_kmr(1152) About Me  (Show Source):
You can put this solution on YOUR website!

y=x^2-17x+30.
x^2-15x -2x +30.
= x(x-15) -2(x-15)
= (x-2)(x-15)

roots are 2 and 15.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
It's rootS, not root!  A quadratic equation 
usually has 2 roots, occasionally just 1 and sometines none at all.

 y = x² - 17x + 30

We factor the right side

Write this:

 y = (x     )(x    )

Think of two whole numbers whose product is 30
and whose sum (since the last sign is +) is 17

They are 15 and 2 because 15·2 = 30 and 15+2 = 17

So fill in 15 and 2 on the right of each parentheses:

 y = (x   15)(x   2)

Since the sign of 17 is -, both signs are -, so we have

 y = (x - 15)(x - 2)

Set each parenthetical expression equal to zero.

   x - 15 = 0   and   x - 2 = 0

        x = 15  and       x = 2

are the two roots of the equation  

x² - 17x + 30 = 0

which are the two x-intercepts of the graph of

y = x² - 17x + 30 = 0

Edwin