Question

If the zeroes of the polynomial $ p{x^2} + qx + r = 0 $ are of the form $ \dfrac{\beta }{{\beta - 1}} $ and \[\dfrac{{\beta + 1}}{\beta }\] . Then the value of $ {(p + q + r)^2} $
(1) $ 4{p^2} - 2pr $
(2) $ {q^2} - 2pr $
(3) $ {q^2} - 4pr $
(4) $ 2{q^2} - pr $

Answer 1

Hint: Use $ p{x^2} + qx + r = (x - \alpha )(x - \beta ) $ and solve for required.
We know that $ p{x^2} + qx + r = (x - \alpha )(x - \beta ) $ , where $ \alpha $ and $ \beta $ are the roots of the given equation. Using that equation, we derive a form, which we will use with another formula of the property of roots which will get us the required value.

Complete step-by-step answer:
 First, we are going to equate the equation to one of the properties of roots which is
 $ p{x^2} + qx + r = (x - \alpha )(x - \beta ) $ , then we get
 $ p{x^2} + qx + r = \left( {x - \dfrac{{\beta + 1}}{\beta }} \right)\left( {x - \dfrac{\beta }{{\beta - 1}}} \right) $
Now, we are going to consider $ x = 1 $ on both RHS and LHS, we get
 $ p + q + r = \left( {1 - \dfrac{{\beta + 1}}{\beta }} \right)\left( {1 - \dfrac{\beta }{{\beta - 1}}} \right) $
We are going to simplify the RHS now, we get
 $ p + q + r = \left( {\dfrac{{\beta - \beta - 1}}{\beta }} \right)\left( {\dfrac{{\beta - 1 - \beta }}{{\beta - 1}}} \right) $
Finally, we will get
 $ p + q + r = \dfrac{1}{{\beta (\beta - 1)}} $
Now, from the above let us consider the RHS and with slight observation with the given roots, it can be written as
 \[\dfrac{1}{{\beta (\beta - 1)}} = \dfrac{{\beta + 1}}{\beta } - \dfrac{\beta }{{\beta - 1}}\]
Now, let us consider the two terms in the RHS as $ \alpha $ and $ \beta $ , which means that
 $ \alpha = \dfrac{{\beta + 1}}{\beta } $ and
 $ \beta = \dfrac{\beta }{{\beta - 1}} $
We know a property of roots which is
 $ \left| {\alpha - \beta } \right| = \sqrt {{b^2} - 4ac} $
 So, on substituting we will get that
 \[\left| {\alpha - \beta } \right| = \sqrt {{q^2} - 4pr} \]
Since, we know that the LHS is equal to $ p + q + r $ , we can substitute. Then, we get
 $ p + q + r = \sqrt {{q^2} - 4pr} $
Since we have to find the value of $ {(p + q + r)^2} $ , we will apply square on both side which we get us to
 \[{\left( {p + q + r} \right)^2} = {\left( {\sqrt {{q^2} - 4pr} } \right)^2}\]
The square root and square will get cancelled, leaving us with the required answer asked in the question
 \[{\left( {p + q + r} \right)^2} = {q^2} - 4pr\]
Which is option (c)
So, the correct answer is “Option C”.

Note: We have to be familiar and strong with the properties of the roots of any given polynomial equation, as without them we cannot proceed any further in this question and it is important to be careful while splitting the denominator in the process.