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..

anyone sir...please help!

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- May 1st 2010, 02:30 AMmastermin346quadratic 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..

anyone sir...please help! - May 1st 2010, 02:41 AMsa-ri-ga-ma
- May 1st 2010, 02:53 AMArchie Meade
Hi mastermin346,

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

the long way to do it is

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

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

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

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

Then twice the second root is the first.

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

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

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

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

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

$\displaystyle 4(3p-4)=36$

$\displaystyle 3p-4=9$

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

The equation is

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

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

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

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