SOLUTION: If you can find this you must be a genius! I've been working on this for half an hour... I would greatly appreciate it! y=6x^2-9x-13 What is the larger intercept? What is the sma

Question 242588: If you can find this you must be a genius! I've been working on this for half an hour... I would greatly appreciate it!
y=6x^2-9x-13
What is the larger intercept? What is the smaller intercept?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
y = 6x^2 -9x - 13

a = 6
b = -9
c = -13

b^2 - 4ac = 81 - (4*6*(-13) = 81 + 312 = 393

393 is not a perfect square so the answer will not be an integer.

quadratic formula is required or completing the square. we'll use quadratic formula.

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

this becomes:

(-(-9) +/- sqrt(393))/12 which becomes:

(9 +/- sqrt(393))/12 which becomes:

x = -.902018967
or:

x = 2.402018967

a graph of your equation looks like this:

I'm not a genius.

I just used the quadratic formula which works every time whether or not you have real roots.

if you can't factor the equation, then use the quadratic formula.

this one could not be factored because the discriminant (b^2-4ac) was not a perfect square.

if b^2-4ac was negative you would not have any real roots because square root of a negative number doesn't yield real roots.

if b^2 - 4ac was equal to 0, you would have only one root.

if b^2 - 4ac was positive, you would have two real roots (your problem fits this).

if b^2 - 4ax was positive and a perfect square, you would have real roots that are integers.

standard form of quadratic equation is:

ax^2 + bx + c

a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant.

your equation was 6x^2 - 9x - 13

that made a = 6, b = -9, c = -13

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