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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 \[{b^2} - 4ac\]>0 | The roots are real and unequal. |

If \[{b^2} - 4ac\]=0 | The roots are real and equal. |

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