Question

The equation \[2{{x}^{2}}+2\left( p+1 \right)x+p=0\], where p is real, always has roots that are
(a) Equal
(b) Equal in magnitude but opposite in sign
(c) Irrational
(d) Real

Answer 1

Hint: In this question, we first need to compare the given quadratic equation with the general form of a quadratic equation given by \[a{{x}^{2}}+bx+c=0\]. Now, substitute the values so obtained in the discriminant formula \[D={{b}^{2}}-4ac\] and find its value. Then from the discriminant value so obtained we can comment on the nature of the roots.

Complete step by step solution:
A quadratic polynomial when equated to zero is called quadratic equation
\[a{{x}^{2}}+bx+c=0\], where \[a\ne 0\]
Roots of a Quadratic Equation:
The values of the variable x which satisfy the quadratic equation are called roots of quadratic equation.
DIRECT FORMULA:
Quadratic equation \[a{{x}^{2}}+bx+c=0\left( a\ne 0 \right)\] has two roots, given by
\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Where, \[D={{b}^{2}}-4ac\] is called the discriminant of the equation
Nature of Roots:
Let the quadratic equation be \[a{{x}^{2}}+bx+c=0\], whose discriminant is D
Now, for \[a{{x}^{2}}+bx+c=0;a,b,c\in R\] and \[a\ne 0\], if
\[D<0\] then the quadratic equation has complex roots
\[D>0\] then the quadratic equation has real and distinct roots
\[D=0\] then the quadratic equation has real and equal roots
Now, from the given quadratic equation in the question we have
\[\Rightarrow 2{{x}^{2}}+2\left( p+1 \right)x+p=0\]
Now, on comparing this with the quadratic equation \[a{{x}^{2}}+bx+c=0\] we have
\[a=2,b=2\left( p+1 \right),c=p\]
Let us now find the discriminant of this quadratic equation using the formula
\[D={{b}^{2}}-4ac\]
Now, on substituting the respective values of a, b, c we get,
\[\Rightarrow D={{\left( 2\left( p+1 \right) \right)}^{2}}-4\times 2\times p\]
Now, this can be further written in the simplified form as
\[\Rightarrow D=4\left( {{p}^{2}}+1+2p \right)-8p\]
Now, this can be further written as
\[\Rightarrow D=4{{p}^{2}}+4+8p-8p\]
Now, on further simplification we get,
\[\Rightarrow D=4\left( {{p}^{2}}+1 \right)\]
Now, this can be further written as
\[\therefore D>0\]
Thus, the given quadratic equation has real and distinct roots
Hence, the correct option is (d).

Note: Instead of finding the discriminant of the given quadratic equation and then simplifying further to get the nature of roots we can also solve this by finding the roots of the given equation using the direct formula which also gives the same answer. This can be done only if we forget the conditions using the value of determinant, else it is better to learn the conditions as it saves time in exams greatly.