Question

If the following equation has no real roots, what are the possible values of \[p\] for \[{x^2} + (3p + 1)x - p = 0\] ?

Answer 1

Hint: Here we go through the properties of the quadratic equation as we know when the roots of the quadratic equation are not real then their discriminant must be less than zero. So we equate the discriminant of this equation less than zero for finding the value of p.
The discriminant of a quadratic equation \[a{x^2} + bx + c = 0\] , where \[a \ne 0\] is given by
\[D = {b^2} - 4ac\]
\[D < 0 \Rightarrow \] Solution of quadratic equation is imaginary
\[D > 0 \Rightarrow \] Solution of quadratic is real and distinct
\[D \ne 0 \Rightarrow \] Solutions are distinct

Complete step-by-step answer:
We know that if in quadratic equation \[a{x^2} + bx + c = 0\] when the two roots are non-real then its discriminant is less than zero i.e. \[{b^2} - 4ac < 0\]
Now in the question the given quadratic equation is \[{x^2} + (3p + 1)x - p = 0\]
By equating it with the general quadratic equation we get \[a = 1,b = 3p + 1,c = - p\]
Now we will calculate its discriminant by formula \[{b^2} - 4ac < 0\]
\[ \Rightarrow {\left( {3p + 1} \right)^2} - 4 \times 1 \times \left( { - p} \right) < 0\]
Using the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] \[ \Rightarrow {(3p)^2} + 2\left( {3p} \right)\left( 1 \right) + {1^2} = 9{p^2} + 6p + 1\]
Substituting this we get
\[ \Rightarrow 9{p^2} + 6p + 1 + 4p < 0\]
\[ \Rightarrow 9{p^2} + 10p + 1 < 0\]
Now take \[10p\] and splitting this into \[9p\] and \[1p\]we get,
\[ \Rightarrow 9{p^2} + 9p + 1p + 1 < 0\]
Taking common terms we get
\[ \Rightarrow 9p\left( {p + 1} \right) + 1\left( {p + 1} \right) < 0\]
\[ \Rightarrow \left( {9p + 1} \right)\left( {p + 1} \right) < 0\]
Therefore p lies between \[ - 1\] and \[\dfrac{{ - 1}}{9}\]
If we break into negative region we get
\[p \in \left( {\dfrac{{ - 1}}{9}, - 1} \right)\]
Therefore,the range of \[p\] is \[p \in \left( {\dfrac{{ - 1}}{9}, - 1} \right)\]
So, the correct answer is “\[p \in \left( {\dfrac{{ - 1}}{9}, - 1} \right)\]”.

Note: If in case we have to find the value of p for two real and unequal roots then we have to do that Discriminant \[{b^2} - 4ac > 0\].If in case we have to find the value of p for imaginary roots then we have to do that Discriminant \[{b^2} - 4ac < 0\].When the Discriminant is negative then we get the pair of complex equations .If one root is given \[r + is\] than other will be the \[r - is\] .