quadratic equation problem

**mastermin346** · May 24th 2010, 09:18 AM

$x^2+p=4x+2$ is three time the other roots.Find the value of $p$ .

please help..i dont know how to do..

**Dinkydoe** · May 24th 2010, 09:33 AM

Can you be a little more specific?

Do you mean that if $x=a$ is a root, then the other root is $x=3a$ ?

Then find the corresponding $p$ ?

**mastermin346** · May 24th 2010, 09:38 AM

Originally Posted by Dinkydoe

Can you be a little more specific?

Do you mean that if $x=a$ is a root, then the other root is $x=3a$ ?

Then find the corresponding $p$ ?

yes.

**papex** · May 24th 2010, 10:09 AM

in that case, rewriting the equation as: x^2-4x+p-2=0 we have -4=4a and p-2=3a^2
thus a=-1 and p=5
the equation becomes x^2-4x+3=0 which has 1 and 3 as roots

**Dinkydoe** · May 24th 2010, 10:12 AM

Ok, write the equation as $x^2-4x+(p-2)= 0$

Suppose that $x^2-4x+(p-2) = (x-a)(x-3a)= x^2-4ax+3a^2$

Now we have $a=1, 3a=3$ and we can solve $p-2=3a^2$ .

**Archie Meade** · May 24th 2010, 10:28 AM

Originally Posted by mastermin346

$x^2+p=4x+2$ One root is three times the other root. Find the value of $p$ .

please help..i dont know how to do..

Hi mastermin346,

Factorising is the fastest way.

$(x-root_1)(x-root_2)=0,\ x(x-root_2)-root_1(x-root_2)=0,$

$x^2-(sum\ of\ roots)x+product\ of\ roots=0$

Here the sum of the roots=4.
One is 3 times the other, so the sum of the roots is 4 times the smaller root.
Smaller root = 4/4.

If you had an expression that was difficult to factorise,
you could proceed as follows.....

The solutions (roots) of a quadratic equation $ax^2+bx+c=0$

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

One root is $\frac{-b+\sqrt{b^2-4ac}}{2a}$

The other root (a quadratic has at most 2 roots) is $\frac{-b-\sqrt{b^2-4ac}}{2a}$

First re-arrange your equation to $ax^2+bx+c=0$

$x^2+p=4x+2\ \Rightarrow\ (1)x^2+(-4)x+(p-2)=0$

$a=1$

$b=-4$

$c=p-2$

If one root is 3 times the other, the bigger root must be the one with the +

$\frac{-b+\sqrt{b^2-4ac}}{2a}=3\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$

The denominator is common, so it is irrelevant.

$-b+\sqrt{b^2-4ac}=3\left(-b-\sqrt{b^2-4ac}\right)$

$4+\sqrt{16-4(p-2)}=3\left(4-\sqrt{16-(p-2)}\right)$

$\sqrt{16-4(p-2)}+3\sqrt{16-4(p-2)}=12-4=8$

$4\sqrt{16-4(p-2)}=8$

This gives $\sqrt{16-4(p-2)}=2$

$16-4(p-2)=4$

$12=4(p-2)$

$3=p-2$

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