Y = 3x^2 - 3x - 6 {the x^2 (x squared) makes it a quadratic formula, and I'm assuming this is what you meant...}

This is derived from:
y = ax^2 + bx + c

So, by using the 'sum and product' rule:

a × c = 3 × (-6) = -18

b = -3

Now, we find the 'sum' and the 'product' of these two numbers, where b is the 'sum' and a × c is the 'product':

The two numbers are: -6 and 3

Proof:

-6 × 3 = -18 {product}

-6 + 3 = -3 {sum}

Now, since a > 1, we divide a from the results

-6/a = -6/3 = -2

3/a = 3/3 = 1

We then implement these numbers into our equation:

(x - 2) × (x + 1) = 0 {derived from 3x^2 - 3x - 6 = 0}

To find x, we make x the subject of 0:

x - 2 = 0

OR

x + 1 = 0

Therefore:

x = 2

OR

x = -1

So the x-intercepts of the quadratic formula (or solutions to equation 3x^2 - 3x -6 = 0, to put it into your words) are 2 and -1.

We can check this by substituting the values for x:

Let's start with x = 2:

y = 3(2)^2 - 3(2) - 6
= 3(4) - 6 - 6
= 12 - 6 - 6
= 0 {so when x = 2, y = 0, which is correct}

For when x = -1:

y = 3(-1)^2 - 3(-1) - 6
= 3(1) + 3 - 6
= 3 + 3 - 6
= 0 {so when x = -1, y = 0, which is correct}

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3 years ago

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