Question

If $f(x)$is a quadratic expression such that $f(1) + f(2) = 0,$and$ - 1$is a root of $f(x) = 0,$then the other root of $f(x) = 0$is:
(A) $ - \dfrac{5}{8}$
(B) $ - \dfrac{8}{5}$
(C ) $\dfrac{5}{8}$
(D) $\dfrac{8}{5}$

Answer 1

Hint: The quadratic equation is in the form of $a{x^2} + bx + c = 0$and if the roots of this equation are $p$and $q$. Then $p + q = \dfrac{{ - b}}{a}$ and $pq = \dfrac{c}{a}$

Complete Step by Step Solution:
Let,$f(x) = a{x^2} + bx + c$then
Put $x = 1$
$ \Rightarrow f(1) = a{(1)^2} + b(1) + c$
$ \Rightarrow f(1) = a + b + c$ . . . . . (1)
Now put $x = 2$
$ \Rightarrow f(2) = a{(2)^2} + b(2) + c$ . . . . . (2)
Given that,
$f(1) + f(2) = 0$
Put the values of $f(1)$ and $f(2)$from equation (1) and (2), in this equation.
$ \Rightarrow a + b + c + 4a + 2b + c = 0$
$ \Rightarrow 5a + 3b + 2c = 0$
Now dividing the equation by $a$, we get
$\dfrac{{5a}}{a} + \dfrac{{3b}}{a} + \dfrac{{2c}}{a} = 0$
$ \Rightarrow 5 + \dfrac{{3b}}{a} + \dfrac{{2c}}{a} = 0$ . . . . . (3)
Now, from the quadratic equation we know that, if $p + q$ are the roots of the equation then,
$p + q = \dfrac{{ - b}}{a}$ and $pq = \dfrac{c}{a}$
One root of the quadratic equation is given to us. Let that root be $p$
$ \Rightarrow p = - 1$
We have, $p + q = \dfrac{{ - b}}{a}$
Put the value of $p $in this equation, we get
$ - 1 + q = \dfrac{{ - b}}{a}$
$(1 - q) = \dfrac{b}{a}$
Rearranging it
$ \Rightarrow \dfrac{b}{a} = 1 - q$ . . . . . (4)
And,$pq = \dfrac{c}{a}$
Put the value of$p$in this equation
We get,
$( - 1) \times q = \dfrac{c}{a}$
$ \Rightarrow \dfrac{c}{a} = - q$ . . . . . . (5)
Put the values from equation (4) and (5) in equation (3), we get
$5 + 3\left( {1 - q} \right) + 2( - q) = 0$
$ \Rightarrow 5 + 3 - 3q + ( - 2q) = 0$
$ \Rightarrow 5 + 3 - 3q - 2q = 0$
$ \Rightarrow 8 - 5q = 0$
$ \Rightarrow 5q = 8$
$ \Rightarrow q = \dfrac{8}{5}$
Therefore, from the above explanation the correct option is [D] $\dfrac{8}{5}$

Note: The quadratic equation $a{x^2} + bx + c = 0$ has an quadratic formula i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
We could have put $x = - 1$ in the formula to find another root by simplifying it. But this method would not have worked in this case, because you would have had only one equation to simplify and three different constants to find the values of. Sometimes you need to think smart to get smart answers.