Obtain the roots of the following quadratic e | vedclass

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(A) The given quadratic equation is $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $

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$x = \frac{q^2}{p^2}, x = -1$

Vedclass · Answer

(A) The given quadratic equation is $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = p^{2}$,$b = (p^{2} - q^{2})$,and $c = -q^{2}$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.First,calculate the discriminant $D = b^{2} - 4ac = (p^{2} - q^{2})^{2} - 4(p^{2})(-q^{2})$.$D = p^{4} - 2p^{2}q^{2} + q^{4} + 4p^{2}q^{2} = p^{4} + 2p^{2}q^{2} + q^{4} = (p^{2} + q^{2})^{2}$.Now,substitute into the formula: $x = \frac{-(p^{2} - q^{2}) \pm \sqrt{(p^{2} + q^{2})^{2}}}{2p^{2}}$.$x = \frac{-p^{2} + q^{2} \pm (p^{2} + q^{2})}{2p^{2}}$.Taking the positive sign: $x = \frac{-p^{2} + q^{2} + p^{2} + q^{2}}{2p^{2}} = \frac{2q^{2}}{2p^{2}} = \frac{q^{2}}{p^{2}}$.Taking the negative sign: $x = \frac{-p^{2} + q^{2} - p^{2} - q^{2}}{2p^{2}} = \frac{-2p^{2}}{2p^{2}} = -1$.Thus,the roots are $x = \frac{q^{2}}{p^{2}}$ and $x = -1$.

Obtain the roots of the following quadratic equation by using the quadratic formula: $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$

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