problem in finding quad. eqn. from the roots

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August 13th 2011, 08:40 PM
sumedh

problem in finding quad. eqn. from the roots

In general, if α(alpha) and ß (beta) are roots of eqn. ax^2 +bx +c=0
then for finding the equation whose roots are α+2 and ß+2 can be done by
addition of roots (α+2+ß+2=-b/a) and product of roots (α+2)(ß+2)=c/a
By solving this we get ax^2 -(4a-b)x + (4a-2b+c)=0

The problem is this that,by replacing x in place of (x-2)in the given equation
ax^2 +bx +c=0 we get the same answer ax^2 -(4a-b)x + (4a-2b+c)=0
but this method ( replacing x by (x-2) ..) is not mentioned anywhere
i am not able to understand this method
please help .
August 13th 2011, 09:49 PM
Prove It

Re: problem in finding quad. eqn. from the roots

You can write every quadratic equation $\displaystyle ax^2 + bx + c = 0$ in terms of its roots, $\displaystyle a\left(x - \alpha\right)\left(x- \beta\right) = 0$ .

The roots of $\displaystyle ax^2 + bx + c = 0$ are $\displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\displaystyle \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$ .

So $\displaystyle \alpha + 2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + 2 = \frac{4a - b + \sqrt{b^2 - 4ac}}{2a}$

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

So for a quadratic which has $\displaystyle \alpha + 2, \beta + 2$ as its roots, the equation is

$\displaystyle \begin{align*} &\phantom{=} A\left[x - \left(\alpha + 2\right)\right]\left[x - \left(\beta + 2\right)\right] \textrm{ where }A\textrm{ is some other parameter} \\ &= A\left[x - \left(\frac{4a - b + \sqrt{b^2 - 4ac}}{2a}\right)\right]\left[x - \left(\frac{4a - b - \sqrt{b^2 - 4ac}}{2a}\right)\right] \end{align*}$

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August 19th 2011, 11:51 PM
sumedh

Re: problem in finding quad. eqn. from the roots

thank you very much i got it