root and equation

**qweiop90** · July 29th 2008, 07:31 AM

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.

answer provided -b/a,c/a, alpha^2,beta^2

**TKHunny** · July 29th 2008, 08:51 AM

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

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 $ax^{2}+bx+c=0$ ?

**qweiop90** · July 29th 2008, 08:56 AM

Originally Posted by TKHunny

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

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 $ax^{2}+bx+c=0$ ?

. i cant be serious? that is the question i copied frm the books?

**TKHunny** · July 29th 2008, 09:29 AM

Why did you change to "alpha" and "beta" later?

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

Can you impose the conditions of the problem statement on those two roots of the equation?

**qweiop90** · July 29th 2008, 09:32 AM

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

**TKHunny** · July 29th 2008, 09:46 AM

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

**qweiop90** · July 29th 2008, 09:56 AM

Originally Posted by TKHunny

Well, $\beta$ has to be the lesser of the two, so $\alpha = 2\beta$ . 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?

**TKHunny** · July 29th 2008, 01:53 PM

$\alpha\;=\;2*\beta\;\implies\;\frac{-b+\sqrt{b^{2}-4ac}}{2a}\;=2*\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

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 $x = \frac{-Horse \pm \sqrt{Horse^{2}-4*Frog*Butterfly}}{2*Frog}$

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