Question

If sin $\alpha$ and cos $\alpha$are roots of the equation $p{x}^2+qx+r=0$ then:

• A. ${\left(p+q\right)}^2=2r$
• B. $p^2-q^2+2pr=0$
• C. $p+q+r=0$
• D. $p-q-r=0$

Answer 1

The correct option is B $p^2-q^2+2pr=0$

$p{x}^2+qx+r=0$

Roots are $\sin \alpha$ and $\cos \alpha$

$\sin \alpha +\cos \alpha =-\frac{q}{p}$

$\sin \alpha \cos \alpha =\frac{r}{p}$

Using Identity ${\sin }^2\alpha +{\cos }^2\alpha =1$

${\sin }^2\alpha +{\cos }^2\alpha =1$

${(\sin \alpha +\cos \alpha )}^2-2\sin \alpha \cos \alpha =1$

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

$q^2-p^2=2rp.$

$\therefore$ $p^2-q^2+2pr=0$