Question

Pqx^2-(p^2+q^2)x+pq=0 solve this problem by using quadratic formula Please don't spam with me

Answer 1

here comes a solution

Answer 2

Answer:

$\frac{p}{q}$ and $\frac{q}{p}$

Step-by-step explanation:

Given:- $pqx^{2} - (p^{2}+ q^{2} )x+pq = 0$

To Find:- Roots using formula of quadratic equation.

Solution:-

$pqx^{2} - (p^{2}+ q^{2} )x+pq = 0$

Here, $a = pq, b = -(p^{2} +q^{2} ), c = pq$

Putting the values in the formula, x = $\frac{-b+\sqrt[]{b^{2}-4ac } }{2a} , \frac{-b-\sqrt[]{b^{2}-4ac } }{2a}$

⇒x =  $\frac{(p^{2} +q^{2} )+\sqrt[]{(-(p^{2} +q^{2}) )^{2}-4(pq)(pq) } }{2(pq)} , \frac{(p^{2} +q^{2} )-\sqrt[]{(-(p^{2} +q^{2}) )^{2}-4(pq)(pq) } }{2(pq)}$

⇒ x = $\frac{(p^{2} +q^{2} )+\sqrt[]{((p^{2} +q^{2}) )^{2}-4(p^{2} q^{2} ) } }{2pq}, \frac{(p^{2} +q^{2} )-\sqrt[]{((p^{2} +q^{2}) )^{2}-4(p^{2} q^{2} ) } }{2pq}$

⇒ x = $\frac{(p^{2} +q^{2} )+\sqrt[]{((p^{2} -q^{2}) )^{2} } }{2pq},\frac{(p^{2} +q^{2} )-\sqrt[]{((p^{2} -q^{2}) )^{2} } }{2pq}$

⇒ x = $\frac{(p^{2} +q^{2} )+{(p^{2} -q^{2}) } }{2pq}, \frac{(p^{2} +q^{2} )-{(p^{2} -q^{2}) } }{2pq}$

⇒ x = $\frac{p^{2} +q^{2} +{p^{2} -q^{2} } }{2pq}, \frac{p^{2} +q^{2} -{p^{2} +q^{2} } }{2pq}$

⇒ x = $\frac{2p^{2} }{2pq}, \frac{2q^{2} }{2pq}$

⇒ x = $\frac{p}{q} , \frac{q}{p}$

Therefore, the roots of the equation $pqx^{2} - (p^{2}+ q^{2} )x+pq = 0$ are $\frac{p}{q}$ and $\frac{q}{p}$.

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