# root and equation

• Jul 29th 2008, 07:31 AM
qweiop90
root and equation
if a and b are the roots of the quadratic equationb ax^2 +bx +c =0 ,express (alpha-2beta)(2alpha-beta) in term if a,b,c . deduce the condition that one root of the equation is twice the other roots.

answer provided :2b^2 = 9ac, (2b^2-9ac)/a^2

2)if,a,b,c is subset of real number, with a is not equal to zero, and the roots of the quadratic equation ax^2+bx+c = 0 are real , show that the roots of ay^2 - (b^2-2ac)y + c^2 = 0 are also real. if the roots of quadratic equation ax^2 + bx +c = 0 states the value of alpha + beta and alpha Xbeta in terms of a,b,c . hence find the roots of the second equation in terms of a and b.

• Jul 29th 2008, 08:51 AM
TKHunny
1) You can't be serious. If a and b are roots of that quadratic, then necessarily $\displaystyle a^{3}+ab+c = 0$ and $\displaystyle b^{2}(a+1) + c = 0$.

2) Are you SURE you mean to use 'a' and 'b' in both places? Perhaps you mean to say that A and B are roots of $\displaystyle ax^{2}+bx+c=0$?
• Jul 29th 2008, 08:56 AM
qweiop90
Quote:

Originally Posted by TKHunny
1) You can't be serious. If a and b are roots of that quadratic, then necessarily $\displaystyle a^{3}+ab+c = 0$ and $\displaystyle b^{2}(a+1) + c = 0$.

2) Are you SURE you mean to use 'a' and 'b' in both places? Perhaps you mean to say that A and B are roots of $\displaystyle ax^{2}+bx+c=0$?

. i cant be serious? that is the question i copied frm the books?
• Jul 29th 2008, 09:29 AM
TKHunny
Why did you change to "alpha" and "beta" later? (Thinking)

We have $\displaystyle \alpha\;=\;\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $\displaystyle \beta\;=\;\frac{-b-\sqrt{b^{2}-4ac}}{2a}$.

Can you impose the conditions of the problem statement on those two roots of the equation?
• Jul 29th 2008, 09:32 AM
qweiop90
oops, i thought you already know . sorry then.

dont get what you mean? currently, i can proved that 2b^2= 9ac. but i couldnt get 1st answer
• Jul 29th 2008, 09:46 AM
TKHunny
Well, $\displaystyle \beta$ has to be the lesser of the two, so $\displaystyle \alpha = 2\beta$. Substitute the larger expressions and see where it leads.
• Jul 29th 2008, 09:56 AM
qweiop90
Quote:

Originally Posted by TKHunny
Well, $\displaystyle \beta$ has to be the lesser of the two, so $\displaystyle \alpha = 2\beta$. Substitute the larger expressions and see where it leads.

alpha = 2 beta?i get what you mean,but can you state the starting, and for the further steps, i will try my hardest effort to solve.

for the question2, which method i should applied with?
• Jul 29th 2008, 01:53 PM
TKHunny
$\displaystyle \alpha\;=\;2*\beta\;\implies\;\frac{-b+\sqrt{b^{2}-4ac}}{2a}\;=2*\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

2) Just use the Quadratic Formula (or Complete the Square if you wish.) It does not matter what the coefficients are, just as long as it LOOKS like a quadratic equation.

Example: If (Frog)x^2 + (Horse)x + Butterfly = 0, then $\displaystyle x = \frac{-Horse \pm \sqrt{Horse^{2}-4*Frog*Butterfly}}{2*Frog}$