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If p, q are real and $p\ne q$ , then the roots of the equation $\left( p-q \right){{x}^{2}}+5\left( p+q \right)x-2\left( p-q \right)=0$ are real and unequal.
(a) True
(b) False

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Last updated date: 25th Apr 2024
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Answer
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Hint: We will first find the discriminant of the quadratic equation given to use using the formula ${{b}^{2}}-4ac$ where $a=\left( p-q \right)$, $b=5\left( p+q \right)$ and $c=-2\left( p-q \right)$. Then, we will see whether the roots obtained are greater than zero or not. After that, we will check the property that is given as If ${{b}^{2}}-4ac > 0$, then the roots of the quadratic equation are real and unequal. If it satisfies then we will get the answer.

Complete step-by-step answer:
Here, we are given with equation $\left( p-q \right){{x}^{2}}+5\left( p+q \right)x-2\left( p-q \right)=0$ and we have to find roots.
So, we will find the discriminant of the equation using the formula ${{b}^{2}}-4ac$ . Here, we have $a=\left( p-q \right)$, $b=5\left( p+q \right)$ and $c=-2\left( p-q \right)$. On putting the values, we get as
Discriminant $={{b}^{2}}-4ac$
$={{\left( 5\left( p+q \right) \right)}^{2}}-4\left( p-q \right)\left( -2\left( p-q \right) \right)$
On further expanding the equation, we get as
$=25{{\left( p+q \right)}^{2}}+8\left( p-q \right)\left( p-q \right)$
$=25{{\left( p+q \right)}^{2}}+8{{\left( p-q \right)}^{2}}>0$
Now, we know the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ which cannot be less than or equal to zero as this is perfect square equation where $a\ne b$ .
We also know the property that If ${{b}^{2}}-4ac>0$ , then the roots of the quadratic equation are real and unequal.
So, here we are given that $p\ne q$ so, if we take any value of p, q and try to solve it by putting the equation we will get the roots real and unequal as roots $25{{\left( p+q \right)}^{2}}$ and $8{{\left( p-q \right)}^{2}}$ are different.
We can also check by taking p as 1, and q as 2. On putting in the equation we get as
$=25{{\left( 1+2 \right)}^{2}}+8{{\left( 1-2 \right)}^{2}}$
On solving we get as
$=\left( 25\times 9 \right)+8=200+8=208$
Thus, 208 is greater than zero. So, we can say that unequal and real.
Thus, we are told in question that rotos of the equation will be real and unequal which is true.

So, the correct answer is “Option A”.

Note: There is also direct property of discriminant which is given as
(1) If ${{b}^{2}}-4ac > 0$ , then the roots of the quadratic equation are real and unequal.
(2) If ${{b}^{2}}-4ac = 0$ , then the roots of the quadratic equation are real and equal.
(3) If ${{b}^{2}}-4ac < 0$ , then roots of quadratic equations are a pair of complex conjugates.
(4) If ${{b}^{2}}-4ac > 0\text{ and perfect square}$ , then the roots of the quadratic equation are real, rational unequal.
So, students should know all these properties then only, it will be easy to identify which type of roots satisfies in this question. So, remember all these properties while solving this type of problem.
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