# Thread: need help ASAP. my exams 2morow

1. ## need help ASAP. my exams 2morow

the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?

2. Originally Posted by outlaw
the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?
Do you mean
$px^2 + 6x + q = 0$
has roots x = -3?

Look at the discriminant of the quadratic:
$D = 6^2 - 4pq = 36 - 4pq$

If the two roots of a quadratic equation are equal, then we know the discriminant is 0:
$36 - 4pq = 0 =\implies q = \frac{9}{p}$

So by the quadratic formula, the solution to $px^2 + 6x + q = 0$ will be:
$x = \frac{-6 \pm D^2}{2p} = -3$

But D = 0, so
$\frac{-6}{2p} = -3$

$-6 = -6p$

$p = 1$

Thus $q = \frac{9}{1} = 9$.

-Dan

3. Originally Posted by outlaw
the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?

the roots of the equation $y = ax^2 + bx + c$ are given by:

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

now for equal roots, we require that $b^2 - 4ac = 0$

now continue

4. I was editing my post in the other thread to note that $b^2-4ac=0$, and then you have two equations of two unknowns p and q to solve for. Follow this thread.

5. If $px^2+6x+q$ has one repeated root; $x=-3$

Then $\implies px^2+6x+q=(x+3)(x+3)$
$\implies px^2+6x+q=x^2+6x+9$

Compare co-effs.

6. Originally Posted by Sean12345
If $px^2+6x+q$ has one repeated root; $x=-3$

Then $\implies px^2+6x+q=(x+3)(x+3)$
$\implies px^2+6x+q=x^2+6x+9$

Compare co-effs.
yup, that's a lot easier. classic, thanks for pointing it out

7. Originally Posted by Sean12345
If $px^2+6x+q$ has one repeated root; $x=-3$

Then $\implies px^2+6x+q=(x+3)(x+3)$
$\implies px^2+6x+q=x^2+6x+9$

Compare co-effs.
This is easier but unfortunately it's also not true in general.

The correct implication is $px^2+6x+q=a(x+3)(x+3)$.

It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.

8. Originally Posted by mr fantastic
This is easier but unfortunately it's also not true in general.

The correct implication is $px^2+6x+q=a(x+3)(x+3)$.

It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.
so true. this caught me in a thread a while back. i realized i needed a constant multiplier when back-substituting. don't remember what thread it was

9. Originally Posted by mr fantastic
This is easier but unfortunately it's also not true in general.

The correct implication is $px^2+6x+q=a(x+3)(x+3)$.

It's only by good luck (a = 1) that this method gave the correct answers. In general you won't be so lucky.
True, i used that method because the 6x shows that a is equal to 1 hence
$px^2+6x+q=1(x+3)(x+3)$