Question

If $p\ne q$ and ${{p}^{2}}=5p-3$ and ${{q}^{2}}=5q-3$ the equation having roots as $\dfrac{p}{q}$and $\dfrac{q}{p}$ is
a) ${{x}^{2}}-19x+3=0$
b) $3{{x}^{2}}-19x-3=0$
c) $3{{x}^{2}}-19x+3=0$
d) $3{{x}^{2}}+19x+3=0$

Answer 1

Let us assume that for a quadratic equation: ${{x}^{2}}+bx+c=0$, we have roots p and q, which forms equation as: ${{p}^{2}}=5p-3$ and ${{q}^{2}}=5q-3$. Therefore, we have a quadratic equation as: ${{x}^{2}}-5x+3=0$. By using sum of roots and product of roots formula, find the value of $\left( p+q \right)\text{ and }pq$. Then, again find the sum of roots and product of roots of $\dfrac{p}{q}$and $\dfrac{q}{p}$. Now, put the values in the standard form of quadratic equation to find the equation having roots as $\dfrac{p}{q}$and $\dfrac{q}{p}$.

Complete step-by-step solution:
As we have assumed that the equation having roots p and q are: ${{x}^{2}}-5x+3=0......(1)$
So, we have:
Sum of roots:
$\begin{aligned} & p+q=-\dfrac{b}{a} \\\ & p+q=\dfrac{5}{1} \\\ & p+q=5......(2) \\\ \end{aligned}$
Product of roots:
$\begin{aligned} & pq=\dfrac{c}{a} \\\ & pq=\dfrac{3}{1} \\\ & pq=3......(3) \\\ \end{aligned}$
For new equation, we have roots as $\dfrac{p}{q}$and $\dfrac{q}{p}$
So, the sum of roots is: $\dfrac{p}{q}+\dfrac{q}{p}=\dfrac{{{p}^{2}}+{{q}^{2}}}{qp}......(4)$
And the product of roots is: $\dfrac{p}{q}\times \dfrac{q}{p}=1......(5)$
We can write equation (4) as:
$\begin{aligned} & \dfrac{p}{q}+\dfrac{q}{p}=\dfrac{{{p}^{2}}+{{q}^{2}}}{qp} \\\ & =\dfrac{{{\left( p+q \right)}^{2}}-2pq}{qp}......(6) \end{aligned}$
Put the value of equation (2) and equation (3) in equation (6), we gat:
$\begin{aligned} & \dfrac{p}{q}+\dfrac{q}{p}=\dfrac{{{\left( 5 \right)}^{2}}-2\times 3}{3} \\\ & =\dfrac{25-6}{3} \\\ & =\dfrac{19}{3}......(6) \end{aligned}$
Now, put the values in the standard form of equation, we get:
$\begin{aligned} & {{x}^{2}}-\left( \dfrac{p}{q}+\dfrac{q}{p} \right)x+\left( \dfrac{p}{q}\times \dfrac{q}{p} \right)=0 \\\ & {{x}^{2}}-\dfrac{19}{3}x+1=0 \\\ & 3{{x}^{2}}-19x+3=0 \\\ \end{aligned}$
Hence the required quadratic equation is $3{{x}^{2}}-19x+3=0$
Therefore, option (c) is the correct answer.

Note: While applying the identity for sum of zeros and product of zeros, always take care of negative sign in the sum of zeros. Some deliberately miss out on the use of the negative signs in the formula and this gives you the wrong value. Also, it was given in the question, that the polynomial is quadratic. For a higher degree of the polynomial, the formula for the sum of zeros and the product of zeros changes accordingly.