Question

solve for x pqx^2-(p^2+q^2)+pq=0
Try to solve the question with quadratic formula

Answer 1

$\underline{\mathcal{GIVEN:}}$

$\textsf{Equation is}$

$\mathsf{pq\,x^2-(p^2+q^2)x+pq=0}$

$\underline{\mathcal{TO\;FIND:}}$

$\textsf{Solution of the given quadratic equation}$

$\underline{\mathcal{SOLUTION:}}$

$\underline{\mathcal{FORMULA\;USED}}$

$\mathsf{The\;roots\;of\;ax^2+bx+c=0\;are}$

$\mathsf{x=\dfrac{-b\pm\,\sqrt{b^2-4ac}}{2a}}$

$\mathsf{Consider}$

$\mathsf{pq\,x^2-(p^2+q^2)x+pq=0}$

$\implies\mathsf{x=\dfrac{-b\pm\,\sqrt{b^2-4ac}}{2a}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)\pm\,\sqrt{(p^2+q^2)^2-4{\times}pq{\times}pq}}{2pq}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)\pm\,\sqrt{p^4+q^4+2p^2q^2-4p^2q^2}}{2pq}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)\pm\,\sqrt{p^4+q^4-2p^2q^2}}{2pq}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)\pm\,\sqrt{(p^2-q^2)^2}}{2pq}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)\pm\,(p^2-q^2)}{2pq}}$

$\implies\mathsf{x=\dfrac{(p^2+q^2)+(p^2-q^2)}{2pq},\;\dfrac{(p^2+q^2)-(p^2-q^2)}{2pq}}$

$\implies\mathsf{x=\dfrac{2p^2}{2pq},\;\dfrac{2q^2}{2pq}}$

$\implies\mathsf{x=\dfrac{p}{q},\;\dfrac{q}{p}}$

$\therefore\mathsf{The\;roots\;are\;\dfrac{p}{q},\;\dfrac{q}{p}}$

$\underline{\mathcal{FIND\;MORE:}}$

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x

https://brainly.in/question/14911682  

Using the quadratic formula, solve the equation a square b square x square - (4b power 4-3apower4)x-12a square b square =0

https://brainly.in/question/5074987  

a(x^2 + 1) = (a^2+ 1) x, a + 0 solve by formula method​

https://brainly.in/question/16238292