First slide

Theory of equations

Question

If the roots of the equation ${ax}^{2} + bx + c = 0$ are of the form $\frac{k + 1}{k}$ and $\frac{k + 2}{k + 1}$ then $(a + b + c)^{2}$ is equal to

Moderate

A

$2 b^{2} - ac$
B

$a^{2}$
C

$b^{2} - 4 ac$
D

$b^{2} - 2 ac$

Solution

We have,
$\frac{k + 1}{k} + \frac{k + 2}{k + 1} = - \frac{b}{a} (1)$
and $\frac{k + 1}{k} \frac{k + 2}{k + 1} = \frac{c}{a}$
$\Rightarrow \frac{k + 2}{k} = \frac{c}{a} or \frac{2}{k} = \frac{c}{a} - 1 = \frac{c - a}{a} or k = \frac{2 a}{c - a} (2)$
Now, eliminate k. Putting the value of k in Eq. (1), we get
$\frac{c + a}{2 a} + \frac{2 c}{c + a} = - \frac{b}{a}$
$\begin{array}{l} or (c + a)^{2} + 4 ac = - 2 b (a + c) \\ or (a + c)^{2} + 2 b (a + c) = - 4 ac \end{array}$
Adding b² to both sides, we have
$(a + b + c)^{2} = b^{2} - 4 ac$

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