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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: 20th Jun 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.

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,
  {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)}} \\
\[ \Rightarrow \dfrac{{p \pm \sqrt {{p^2} - 4q} }}{{ - 2p}}\]
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 \[{b^2} - 4ac\]>0The roots are real and unequal.
If \[{b^2} - 4ac\]=0The roots are real and equal.