1. ## value of m

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

can any body help me on this ?

thanks

2. ## Re: value of m

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

Do you see the sum of the roots?

3. ## Re: value of m

Originally Posted by Plato
$\displaystyle (x-q)(x-p)=x^2-(p+q)x+p\cdot q=0$.

Do you see the sum of the roots?
so therefore p+q = 6 ... how is it possible to get p x q = 5? ...

4. ## Re: value of m

Originally Posted by rcs
how is it possible to get p x q = 5? ...

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

5. ## Re: value of m

Hello, rcs!

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

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

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

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

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

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

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

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

Check:

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

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

6. ## Re: value of m

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