Question

find the quadratic equation ,whose roots are p+q/p and p+q/q also find the nature of the rootswhen p=2 and q=3​

Answer 1

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Answer 2

The quadratic equation is $6x^2-25x+25=0$

Step-by-step explanation:

Given the roots are $\frac{p+q}{p}$ and $\frac{p+q}{q}$

Also given p=2 and q=3

To find the nature of the roots when p=2 and q=3 and its quadratic equation:

Let $\alpha$ and $\beta$ be the two given roots $\frac{p+q}{p}$ and $\frac{p+q}{q}$ respectively

Put p=2 and q=3 we have the roots $\alpha=\frac{2+3}{2}$ and $\beta=\frac{2+3}{3}$

$\alpha=\frac{5}{2}$ and $\beta=\frac{5}{3}$

• Now sum the roots we get

$\alpha+\beta=\frac{5}{2}+\frac{5}{3}$

$=\frac{5(3)+5(2)}{6}$

$=\frac{15+10}{6}$

$=\frac{25}{6}$

$\alpha+\beta=\frac{25}{6}$

• Product of the roots

$\alpha\times \beta=\frac{5}{2}\times \frac{5}{3}$

$=\frac{25}{6}$

$\alpha.\beta=\frac{25}{6}$

For quadratic equation we have $ax^2+bx+c=0$

sum of the roots $\alpha+\beta=-\frac{b}{a}$

$=\frac{25}{6}$

Comparing both the equations we have a=6 and b=25

Product of the roots $\alpha\beta=\frac{c}{a}$

$=\frac{25}{6}$

Comparing both the equations we have a=6 and c=25

$ax^2-(sum of the roots)x+(product of the roots)=0$

Therefore the quadratic equation is $6x^2-25x+25=0$

Now Discriminant $D=b^2-4ac$

$=(25)^2-4(6)(25)$

$=625-600$

$=25$

Therefore D=25>0

Therefore if D>0 then the given roots are real and unequal