Question

If the roots of ${x^3} - 7{x^2} + 14x + k = 0$ are in G.P. then k=?
A.-8
B.8
C.7
D.-7

Answer 1

Use relation between roots of cubic equation. Sum of the roots is equal to the negative of the ratio of coefficients of second term to first term. The product of the roots is equal to the negative of ratio of coefficients of last term to first term. The sum of products of roots is equal to the ratio of coefficients of third term to first term.

Complete step-by-step answer:
Given equation ${x^3} - 7{x^2} + 14x + k = 0$ is a cubic equation of the form $a{x^3} + b{x^2} + cx + d = 0$.
Let’s say p, q and r are the roots of the equation.
Using relation between the roots ,
Sum of the roots is equal to the negative of the ratio of coefficients of second term to first term.

$$\Rightarrow p + q + r = - \dfrac{{\left( { - 7} \right)}}{1} \\\ \Rightarrow p + q + r = 7 \\\$$

The product of the roots is equal to the negative of ratio of coefficients of last term to first term.

$$\Rightarrow pqr = - \dfrac{k}{1} \\\ \Rightarrow pqr = - k \\\$$

The sum of products of roots is equal to the ratio of coefficients of third term to first term.
$$
\Rightarrow pq \times qr \times rp = \dfrac{{14}}{1} \\
\Rightarrow pq \times qr \times rp = 14 \\

$$But according to the given condition the roots are in geometric progression. So ,$$

\Rightarrow {q^2} = pr \\
\Rightarrow pqr = - k \\
\Rightarrow q.{q^2} = - k \\
\Rightarrow {q^3} = - k \\

Now taking cube root on both sides, $$ \Rightarrow q = {\left( { - k} \right)^{\dfrac{1}{3}}}$$ Now putting this value in given equation

\Rightarrow {\left( {{{\left( { - k} \right)}^{\dfrac{1}{3}}}} \right)^3} - 7{\left( {{{\left( { - k} \right)}^{\dfrac{1}{3}}}} \right)^2} + 14{\left( { - k} \right)^{\dfrac{1}{3}}} + k = 0 \\
\Rightarrow - k + 7{\left( k \right)^{\dfrac{2}{3}}} - 14{\left( k \right)^{\dfrac{1}{3}}} + k = 0 \\
\Rightarrow 7{\left( k \right)^{\dfrac{2}{3}}} - 14{\left( k \right)^{\dfrac{1}{3}}} = 0 \\
\Rightarrow 7{\left( k \right)^{\dfrac{2}{3}}} = 14{\left( k \right)^{\dfrac{1}{3}}} \\
\Rightarrow {\left( k \right)^{\dfrac{2}{3}}} = 2{\left( k \right)^{\dfrac{1}{3}}} \\
\Rightarrow {\left( k \right)^{\dfrac{2}{3} - \dfrac{1}{3}}} = 2 \\
\Rightarrow {\left( k \right)^{\dfrac{1}{3}}} = 2 \\

On cubing both sides, $$ \Rightarrow k = 8$$ Hence value of k=8. **Thus option B is the correct answer.** **Note:** Using the relation between roots of a cubic equation, we can easily get the value of k. No need to find the actual roots. Just use the condition applied on roots in the given question.