Question
The difference between the roots of the equation $x^2-13 x+\mathrm{k}=0$ is 7 find $\mathrm{k}$.
✓
Answer
Comparing $x^2-13 x+\mathrm{k}=0$ with $a x^2+b x+c=0$
$a=1, b=-13, c=\mathrm{k}$
Let $\alpha$ and $\beta$ be the roots of the equation.
$\alpha+\beta=-\frac{b}{a}=-\frac{(-13)}{1}=13 \ldots$
But
$\alpha-\beta =7 \ldots \ldots \ldots \text { (given) (II) }$
$2 \alpha =20 \ldots \text { (adding (I) and (II)) }$
$\therefore \alpha =10$
$\therefore 10 +\beta=13 \ldots(\text { from (I)) }$
$\therefore \beta =13-10$
$\therefore \beta =3 $
But $\alpha \times \beta=\frac{c}{a}$
$ \therefore 10 \times 3=\frac{k}{1}$
$\therefore \mathrm{k}=30 $
$a=1, b=-13, c=\mathrm{k}$
Let $\alpha$ and $\beta$ be the roots of the equation.
$\alpha+\beta=-\frac{b}{a}=-\frac{(-13)}{1}=13 \ldots$
But
$\alpha-\beta =7 \ldots \ldots \ldots \text { (given) (II) }$
$2 \alpha =20 \ldots \text { (adding (I) and (II)) }$
$\therefore \alpha =10$
$\therefore 10 +\beta=13 \ldots(\text { from (I)) }$
$\therefore \beta =13-10$
$\therefore \beta =3 $
But $\alpha \times \beta=\frac{c}{a}$
$ \therefore 10 \times 3=\frac{k}{1}$
$\therefore \mathrm{k}=30 $
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