Question

If the roots of the equation $a{x}^2+bx+c=0$ are of the form $\frac{(k+1)}{k}$ and$\frac{(k+2)}{(k+1)}$ then ${\left(a+b+c\right)}^2=?$

• A. $b^2-4ac$
• B. $b^2-2ac$
• C. $2b^2-ac$
• D. $\sum _{}{a}^2$

Answer 1

The correct option is A

$b^2-4ac$

Step 1: Use the concept of Sum and Product of roots of a Quadratic equation

We know that for a quadratic equation $a{x}^2+bx+c=0$,

The sum of the roots $=\frac{-b}{a}$

The product of the roots $=\frac{c}{a}$

Since $\frac{(k+1)}{k}$ and $\frac{(k+2)}{(k+1)}$ are roots,

$\therefore \frac{[k+1]}{k}+\frac{[k+2]}{[k+1]}=-\frac{b}{a}$ $....\left(1\right)$

And $\frac{k+2}{k}=\frac{c}{a}$ $....\left(2\right)$

Step 2: Eliminate $k$ from the above pair of equations

Simplifying $\left(2\right)$, we get

$\frac{k+2}{k}=\frac{c}{a}$

$\Rightarrow 1+\frac{2}{k}=\frac{c}{a}$

$\Rightarrow k=\frac{2a}{(c-a)}$ $....\left(3\right)$

Simplifying $\left(1\right)$, we get

$\frac{[k+1]}{k}+\frac{[k+2]}{[k+1]}=-\frac{b}{a}$

$$\begin{gathered} \Rightarrow 1+\frac{1}{k}+1+\frac{1}{k+1}=-\frac{b}{a} \\ \Rightarrow 2+\frac{1}{k}+\frac{1}{k+1}=-\frac{b}{a} \end{gathered}$$

Substituting the value of $k$ from $\left(3\right)$, we get

$2+\frac{c-a}{2a}+\frac{c-a}{c+a}=-\frac{b}{a}$

Step 3: Simplify the above equation

$\frac{[4a(a+c)+(c-a\left)\right(c+a)+2a(c-a\left)\right]}{[2a(a+c\left)\right]}=-\frac{b}{a}$'

$\begin{array}{rcl}4a^2+4ac+c^2-a^2+2ac-2a^2& =& -2b\left(a+c\right)\\ & \Rightarrow & a^2+c^2+6ac=-2ba-2bc\\ & \Rightarrow & a^2+c^2=-6ac-2ba-2bc\\ & \Rightarrow & a^2+c^2+b^2=-6ac-2ba-2bc+b^2\\ & \Rightarrow & a^2+b^2+c^2+2ab+2bc+2ca=b^2-4ac\\ & \Rightarrow & {\left(a+b+c\right)}^2=b^2-4ac\end{array}$

Hence, the value of $(a+b+c{)}^2=b^2-4ac$

Thus, the correct option is A.