1. ## quadratic equation problem

Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

a)value of p

b) root of the quadratic equation..

2. Originally Posted by mastermin346
Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

a)value of p

b) root of the quadratic equation..

If α and β are the roots of a quadratic equation ax^2 + bx + c = 0, then
the product of the roots α*β = c/a. and sum of the roots (α + β) = -(b/a)
Using sum of the roots equation, find α. Using the product of roots equation, find p.

3. Originally Posted by mastermin346
Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

a)value of p

b) root of the quadratic equation..

Hi mastermin346,

sa-ri-ga-ma's method is a fast way...

the long way to do it is

the equation in $ax^2+bx+c$ form is $8x^2-18x+(3p-4)=0$

One root is $\frac{18+\sqrt{18^2-32(3p-4)}}{16}$

The other root is $\frac{18-\sqrt{18^2-32(3p-4)}}{16}$

The larger root must be the first one if the roots are real.

Then twice the second root is the first.

$2\left(18-\sqrt{324-32(3p-4)}\right)=18+\sqrt{324-32(3p-4)}$

$18=3\sqrt{324-32(3p-4)}$

$\sqrt{324-32(3p-4)}=6$

$324-32(3p-4)=36$

$32(3p-4)=324-36=288$

$4(3p-4)=36$

$3p-4=9$

$p=\frac{13}{3}$

The equation is

$8x^2-18x+13-4=0$

$8x^2-18x+9=0$

Either find the roots of this or substitute "p" into

$\frac{18\pm\sqrt{324-32(9)}}{16}$