Question

The equation $$x^2-px+q=0$$ where $$p,q\in R$$ has no real roots if
A.$$p^2>4q$$
B.$$p^2<4q$$
C.$$p^2=4q$$
D.None of these

Answer 1

Hint: Given equation $$x^2-px+q=0$$ is of the form $$a{x}^2+bx+c=0$$ . This is a quadratic equation. The nature of the roots of a quadratic equation depends on the term $$b^2-4ac$$ . So let’s check it with the given equation.

Complete step-by-step answer:
Now the given equation is $$x^2-px+q=0$$. Comparing this with the general quadratic equation $$a{x}^2+bx+c=0$$ we get a=1, b=-p and c=q. Also it is given that $$p,q\in R$$.
The term that decides the nature of the roots of the equation is,
$$\begin{gathered} b^2-4ac\Rightarrow {\left(-p\right)}^2-4\times 1\times q \\ \Rightarrow p^2-4q \end{gathered}$$
Now if $$p^2-4q<0$$ then the given equation has no real roots because, roots of the equation is given by,
$$\begin{gathered} \Rightarrow {\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}} \\ \Rightarrow {\displaystyle \frac{-\left(-p\right)\pm \sqrt{{\left(-p\right)}^2-4\times 1\times q}}{2\left(-p\right)}} \end{gathered}$$
$$\Rightarrow {\displaystyle \frac{p\pm \sqrt{p^2-4q}}{-2p}}$$
Now if the term in the square root is negative or less than zero then the roots so obtained are imaginary. Thus the equation has no real roots then.

Note: We will also have a look on other conditions of the nature of the roots .

If $$b^2-4ac$$>0	The roots are real and unequal.
If $$b^2-4ac$$=0	The roots are real and equal.