Question

solve the following quadratic equation by applying the quadratic formula p square x square + b square minus Q Square x minus Q square equals to zero

Answer 1

Answer:

x= -1

and

x= q2/p2

thank you

Answer 2

Using quadratic formula solve the given quadratic equation: $p^2x^2+(p^2-q^2)x-q^2=0$

Explanation:  

1. For a given quadratic equation in 'x' where 'a' 'b' and 'c' are real numbers,    $ax^2+bx+c=0$
2. We define the roots of the equation $(\alpha,\ \beta)$ using the quadratic formula as,                                                                                                    $(\alpha,\ \beta)=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$       ----(a)
3. We have equation,   $p^2x^2+(p^2-q^2)x-q^2=0$  having roots $(\alpha,\ \beta)$.
4. hence,  $a=p^2,\ \ b=p^2-q^2,\ \ c=-q^2$  
5. putting these values in (a) we get,                                                         [tex](\alpha,\ \beta)=\frac{-(p^2-q^2)\pm\sqrt{(p^2-q^2)^2+4p^2q^2} }{2p^2}\\ (\alpha,\ \beta)=\frac{-p^2+q^2\pm\sqrt{p^4+q^4-2p^2q^2+4p^2q^2} }{2p^2}\\ \\ (\alpha,\ \beta)=\frac{-p^2+q^2\pm\sqrt{(p^2+q^2)^2} }{2p^2}\\\\ (\alpha,\ \beta)=\frac{-p^2+q^2\pm(p^2+q^2) }{2p^2}\\\\ ->\alpha=\frac{-p^2+q^2+p^2+q^2 }{2p^2}\ \ \ \ \ \ \beta= \frac{-p^2+q^2-p^2-q^2 }{2p^2} \\ ->\alpha=\frac{q^2}{p^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta=-1 [/tex]      
6. hence roots of the given equation are $(\frac{q^2}{p^2} ,-1)$