qu-quadratic eqn.

**magaski** · January 17th 2010, 09:35 AM

Find the value of p so that the equation x^2+10x+21=0 and x^2+9x+p=0 may have a common root. Find also the equation formed by the other root.

**Dinkydoe** · January 17th 2010, 09:53 AM

Observe that $f(x) = x^2+10x+21 = (x+3)(x+7)$ . Hence roots of $f(x) = 0$ are $x=-3$ and $x=-7$ .

Observe that roots of $g(x) = x^2+9x+p =0$ are given by $x_0 = \frac{-9\pm \sqrt{81-4p}}{2}$ .

Thus if you find the values of $p$ where $x_0 = -3$ or $x_0 = -7$ you're done.

**Dinkydoe** · January 17th 2010, 10:06 AM

Sorry. This can be done easier ;p.

Since you must find p such that $(x-a)(x+3) =x^2+(3-a)x-3a = x^2+9x+p$ . Observe the only possibility is $a = -\frac{p}{3}$ .

Thus now we must find p such that: $x^2+(3+\frac{p}{3})x+ p= x^2+9x+p$ . This gives $p = 18$

Now you can do the same trick for $f(x) = (x-a)(x+7)$ to find another value of p

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