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# 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

Last updated date: 11th Sep 2024
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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.

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 $p,q \in R$.
${b^2} - 4ac \Rightarrow {\left( { - p} \right)^2} - 4 \times 1 \times q \\ \Rightarrow {p^2} - 4q \\$
Now if ${p^2} - 4q < 0$ then the given equation has no real roots because, roots of the equation is given by,
$\Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\ \Rightarrow \dfrac{{ - \left( { - p} \right) \pm \sqrt {{{\left( { - p} \right)}^2} - 4 \times 1 \times q} }}{{2\left( { - p} \right)}} \\$
$\Rightarrow \dfrac{{p \pm \sqrt {{p^2} - 4q} }}{{ - 2p}}$
 If ${b^2} - 4ac$>0 The roots are real and unequal. If ${b^2} - 4ac$=0 The roots are real and equal.