#### mastermin346

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

#### Dinkydoe

Can you be a little more specific?

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

Then find the corresponding $$\displaystyle p$$?

#### mastermin346

Can you be a little more specific?

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

Then find the corresponding $$\displaystyle p$$?

yes.

#### papex

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

• mastermin346

#### Dinkydoe

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

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

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

• mastermin346

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

Hi mastermin346,

Factorising is the fastest way.

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

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

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 $$\displaystyle ax^2+bx+c=0$$

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

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

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

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

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

$$\displaystyle a=1$$

$$\displaystyle b=-4$$

$$\displaystyle c=p-2$$

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

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

The denominator is common, so it is irrelevant.

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

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

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

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

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

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

$$\displaystyle 12=4(p-2)$$

$$\displaystyle 3=p-2$$

• mastermin346