Question

If the roots of the equation $a{{x}^{2}}+bx+c=0$ are in the form $\dfrac{\left( k+1 \right)}{k}$ and $\dfrac{k+2}{k+1}$ , then ${{\left( a+b+c \right)}^{2}}$ is equal to
(a) $2{{b}^{2}}-ac$
(b) ${{a}^{2}}$
(c) ${{b}^{2}}-4ac$
(d) ${{b}^{2}}-2ac$

Answer 1

In order to solve this problem, we need to find out the property of the quadratic equation. The property states as, in $a{{x}^{2}}+bx+c=0$ , the sum of the roots is given by $\dfrac{-b}{a}$ . Also, in $a{{x}^{2}}+bx+c=0$ , the product of the roots is given by $\dfrac{c}{a}$ . Now we can eliminate k and get the equation for ${{\left( a+b+c \right)}^{2}}$ . For simplification we must know the following identities, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$ .

Complete step-by-step solution:
We have a quadratic equation $a{{x}^{2}}+bx+c=0$ and we also know the root of this equation.
The roots of this quadratic equation are $\dfrac{\left( k+1 \right)}{k}$ and $\dfrac{k+2}{k+1}$.
There is a relation between the roots of the quadratic equation and the coefficients of the same equation.
It is given as follows,
In $a{{x}^{2}}+bx+c=0$ , the sum of the roots is given by $\dfrac{-b}{a}$ .
Also, in $a{{x}^{2}}+bx+c=0$ , the product of the roots is given by $\dfrac{c}{a}$ .
According to the first condition we get,
$\dfrac{k+1}{k}+\dfrac{k+2}{k+1}=\dfrac{-b}{a}...............(i)$
And according to the second equation we get,
$\left( \dfrac{k+1}{k} \right)\left( \dfrac{k+2}{k+1} \right)=\dfrac{c}{a}...................(ii)$
Now our aim is to eliminate k by using (i) and (ii).
Let's start by solving (ii), we get,
$\begin{aligned} & \left( \dfrac{k+1}{k} \right)\left( \dfrac{k+2}{k+1} \right)=\dfrac{c}{a} \\\ & \dfrac{k+2}{k}=\dfrac{c}{a} \\\ & 1+\dfrac{2}{k}=\dfrac{c}{a} \\\ & \dfrac{2}{k}=\dfrac{c}{a}-1 \\\ & \dfrac{2}{k}=\dfrac{c-a}{a} \\\ & \dfrac{k}{2}=\dfrac{a}{c-a} \\\ & k=\dfrac{2a}{c-a} \\\ \end{aligned}$
Now we have to substitute the value of k in equation (ii)
$\begin{aligned} & \dfrac{k+1}{k}+\dfrac{k+2}{k+1}=\dfrac{-b}{a} \\\ & \dfrac{\dfrac{2a}{c-a}+1}{\dfrac{2a}{c-a}}+\dfrac{\dfrac{2a}{c-a}+2}{\dfrac{2a}{c-a}+1}=\dfrac{-b}{a} \\\ & \dfrac{\dfrac{2a+c-a}{c-a}}{\dfrac{2a}{c-a}}+\dfrac{\dfrac{2a+2c-2a}{c-a}}{\dfrac{2a+c-a}{c-a}}=\dfrac{-b}{a} \\\ & \dfrac{a+c}{2a}+\dfrac{2c}{a+c}=\dfrac{-b}{a} \\\ \end{aligned}$
We need to solve this further by cross multiplying,
$\begin{aligned} & \dfrac{a+c}{2a}+\dfrac{2c}{a+c}=\dfrac{-b}{a} \\\ & \dfrac{{{\left( a+c \right)}^{2}}+4ac}{2a\left( a+c \right)}=\dfrac{-b}{a} \\\ & {{\left( a+c \right)}^{2}}+4ac=-b\left( 2a+2c \right) \\\ \end{aligned}$
We must know the property, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .
Therefore, substituting we get,
$\begin{aligned} & {{\left( a+c \right)}^{2}}+4ac=-b\left( 2a+2c \right) \\\ & {{a}^{2}}+{{c}^{2}}+2ac+4ac=-2ba-2bc \\\ & {{a}^{2}}+{{c}^{2}}+2ab+2bc+6ac=0 \\\ \end{aligned}$
The formula for ${{\left( a+b+c \right)}^{2}}$ is given by ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$ .
Therefore, adding ${{b}^{2}}$ we get,
$\begin{aligned} & {{a}^{2}}+{{c}^{2}}+2ab+2bc+6ac=0 \\\ & {{a}^{2}}+{{c}^{2}}+{{b}^{2}}+2ab+2bc+6ac={{b}^{2}} \\\ & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac+4ac={{b}^{2}} \\\ & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac+4ac={{b}^{2}} \\\ & {{\left( a+b+c \right)}^{2}}={{b}^{2}}-4ac \\\ \end{aligned}$

Hence, the correct option is (c).

Note: In this problem, the calculation can get very tricky. While eliminating k we can also use equation (i) to find the equation of k and then substitute in equation (ii). We will get the same answer. Also, we must be aware of the formula for ${{\left( a+b+c \right)}^{2}}$ and add ${{b}^{2}}$ on both sides.