value of m

**rcs** · January 21st 2013, 05:09 AM

solve for x^2 - mx - 5 whose sum of roots is 6. find m.

can any body help me on this ?

thanks

**Plato** · January 21st 2013, 05:28 AM

Originally Posted by rcs

solve for x^2 - mx - 5 whose sum of roots is 6. find m.
can any body help me on this ?

$(x-q)(x-p)=x^2-(p+q)x+p\cdot q=0$ .

Do you see the sum of the roots?

**rcs** · January 21st 2013, 06:35 AM

Originally Posted by Plato

$(x-q)(x-p)=x^2-(p+q)x+p\cdot q=0$ .

Do you see the sum of the roots?

so therefore p+q = 6 ... how is it possible to get p x q = 5? ...

**Plato** · January 21st 2013, 01:59 PM

Originally Posted by rcs

how is it possible to get p x q = 5? ...

It is not possible because $p\cdot q=-5~.$

**Soroban** · January 21st 2013, 02:49 PM

Hello, rcs!

$\text{The sum of roots of }\,x^2 - mx - 5\:=\:0\,\text{ is 6. Find }m.$

Suppose the two roots are $p$ and $q$ .

Vieta's formulas tell us: . $p+q \:=\:m,\quad pq \:=\:-5$

. . Therefore: . $m \,=\,6$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The equation becomes: . $x^2 - 6x - 5 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{6 \pm\sqrt{36+20}}{2} \;=\;3 \pm\sqrt{14}$

Hence: . $\begin{Bmatrix}p &=& 3 + \sqrt{14} \\ q &=& 3 - \sqrt{14}\end{Bmatrix}$

Check:

$p+q \;=\;(3+\sqrt{14}) + (3 - \sqrt{14}) \;=\;6\;\;\checkmark$

$pq \;=\;(3 + \sqrt{14})(3-\sqrt{14}) \;=\;9 - 14 \;=\;-5\;\;\checkmark$

**rcs** · January 21st 2013, 09:32 PM

Wow i got it. Thanks.... My god this is magic ... Thanks Sir Soroban

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