1. ## Algeb

If $x^2+abx+c=0 \mbox\ {and} \ x^2+acx+b=0$ quadratic equations have a common root.prove b=c or the other roots are the roots of $a(b+c)x^2+(b+c)-abc=0$

2. ## Re: Algeb

Originally Posted by srirahulan
If $x^2+abx+c=0 \mbox\ {and} \ x^2+acx+b=0$ quadratic equations have a common root.prove b=c or the other roots are the roots of $a(b+c)x^2+(b+c)-abc=0$
The roots of the first equation are

\displaystyle \begin{align*} x &= \frac{-ab \pm \sqrt{(ab)^2 - 4(1)(c)}}{2(1)} \\ &= \frac{-ab \pm \sqrt{a^2b^2 - 4c}}{2} \end{align*}

and the roots of the second equation are

\displaystyle \begin{align*} x &= \frac{-ac \pm \sqrt{(ac)^2 - 4(1)(b)}}{2(1)} \\ &= \frac{-ac \pm \sqrt{a^2c^2 - 4b}}{2} \end{align*}

Since they share a root, there is some value such that

\displaystyle \begin{align*} \frac{-ab \pm \sqrt{a^2b^2 - 4c}}{2} &= \frac{-ac \pm \sqrt{a^2c^2 - 4b}}{2} \\ -ab \pm \sqrt{a^2b^2 - 4c} &= -ac \pm \sqrt{a^2c^2 - 4b} \end{align*}

If \displaystyle \begin{align*} b = c \end{align*}, we have

\displaystyle \begin{align*} LHS &= -ab \pm \sqrt{a^2b^2 - 4c} \\ &= -ab \pm \sqrt{a^2b^2 - 4b} \\ \\ RHS &= -ac \pm \sqrt{a^2c^2 - 4b} \\ &= -ab \pm \sqrt{a^2b^2 - 4b} \\ &= LHS \end{align*}

So that means for the two equations to share a root, it's possible that \displaystyle \begin{align*} b = c \end{align*}.

3. ## Re: Algeb

I can't understand you statements.pls reply,

4. ## Re: Algeb

Originally Posted by srirahulan
....are the roots of $a(b+c)x^2+(b+c)-abc=0$
k = [abc - (b+c)] / [a(b+c)]
x = +- SQRT(k)

5. ## Re: Algeb

Originally Posted by srirahulan
If $x^2+abx+c=0 \mbox\ {and} \ x^2+acx+b=0$ quadratic equations have a common root.prove b=c or the other roots are the roots of $a(b+c)x^2+(b+c)-abc=0$
x^2 + abx + c = 0 [1]
x^2 + acx + b = 0 [2]

Roots of [1] = u,v
Roots of [2] = u,w
Then:
ab = u + v
c = uv : v = c/u
ab = u + c/u : a = (u + c/u) / b [3]

Similarly: a = (u + b/u) / c [4]

[3][4]: (u + c/u) / b = (u + b/u) / c
Simplify:
bu^2 + b^2 = cu^2 + c^2
b = c