Question

Using quadratic formula,solve the following quadratic equation for x:
p²x²+(p²-q²)x-q²=0​

Answer 1

Roots of quadratic equation for x are $\frac{q^{2} }{^{p2} }$ and -1.

Step-by-step explanation:

We are given ,

p²x²+(p²-q²)x-q²=0​

and we have to find the roots of given equation with respect to x.

• Formula used,

We use Quadratic formula for solving the question. Let us take a quadratic equation to understand the quadratic formula.

For example, we have a quadratic equation,

$a*x^{2} +b*x+c=0$        ---- (1)

for solving this equation we have to find Discriminant first which is denoted by D and has formula ,

$D = b^{2} -4*a*c$    

( a and b are the coefficients of $x^{2}$ and x respectively and c is the constant in quadratic equation.)

Now after Discriminant the next step is to find roots of quadratic equation,

$x = \frac{-b\pm\sqrt{D} }{2*a}$  

by this we get two roots of x which re,

$x1=\frac{-b+\sqrt{D} }{2*a}$    and   $x2 = \frac{-b-\sqrt{D} }{2*a}$  

• Find roots of given quadratic equation w.r.t x

we have , p²x²+(p²-q²)x-q²=0​

If we compare this equation with the standard equation (1) then we get values of a , b and c are

$a=p^{2}$      ,    $b = p^{2} -q^{2}$   and $c = -q^{2}$

The value of Discriminant is ,

$D = b^{2} -4*a*c$  

    $= (p^{2} -q^{2} )^{2} - 4*p^{2} (-q)^{2}$

    $= (p^{2} )^{2} +(q^{2} )^{2} -2p^{2} q^{2} +4p^{2} q^{2}$

    $= p^{4} +q^{4} +2p^{2} q^{2}$

   $=(p^{2} +q^{2} )^{2}$

Now the x,

$x = \frac{-b\pm\sqrt{D} }{2*a}$  

  $= \frac{-(p^{2} -q^{2} )\pm\sqrt{(p^{2} +q^{2} ){2} }}{2*p^{2} }$

now $x1$ and $x2$ are,

$x1=\frac{-b+\sqrt{D} }{2*a}$                                     and           $x2 = \frac{-b-\sqrt{D} }{2*a}$  

$x1= \frac{-p^{2} +q^{2} +{(p^{2} +q^{2} ) }}{2*p^{2} }$                                          $x2= \frac{-p^{2} +q^{2} -{(p^{2} +q^{2} ) }}{2*p^{2} }$  

$x1= \frac{-p^{2}+q^{2} +p^{2} +q^{2} }{2*p^{2} }$                                             $x2= \frac{-p^{2}+q^{2} -p^{2} -q^{2} }{2*p^{2} }$

$x1= \frac{2*q^{2} }{2*p^{2} }$                                                            $x2=\frac{-2*p^{2} }{2*p^{2} }$

$x1= \frac{q^{2} }{p^{2} }$                                                               $x2= -1$

 we get $x1 ,x2$ as

$x1=\frac{q^{2} }{p^{2} }$                  and     $x2=-1$

Roots of Quadratic equation are $\frac{q^{2} }{^{p2} }$  and -1.