Question

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

Answer 1

Hint: Firstly understand the information given in the question. Start with assuming the function $f\left( x \right)$ as $\left( {x - \alpha } \right)\left( {x - \beta } \right)$ where $\alpha $ and $\beta $ are the roots of the equation. Now put $\alpha = - 1$ in the function as it is given. Now use the function definition to satisfy the equation $f\left( 1 \right) + f\left( 2 \right) = 0$ and solve it to find the only unknown value of $\beta $ .

Complete step-by-step solution:
Let’s first try to analyse the question properly. Here we have a quadratic polynomial $f\left( x \right)$ which satisfies the equation $f\left( 1 \right) + f\left( 2 \right) = 0$ and it has a root $ - 1$ . With this information, we need to find the other root for $f\left( x \right) = 0$
As we know that, the general form of a quadratic equation is $a{x^2} + bx + c = 0$ and the part on the left side is called the general form of a quadratic expression. And the quadratic expression in variable $x$ and with roots as $\alpha $ and $\beta $ can be written as $\left( {x - \alpha } \right)\left( {x - \beta } \right)$
Therefore, let’s assume $f\left( x \right) = \left( {x - \alpha } \right)\left( {x - \beta } \right) = {x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta $
But it is already given that the equation $f\left( x \right) = 0$ has a root $ - 1$
Hence, we can write the function as \[f\left( x \right) = \left( {x + 1} \right)\left( {x - \beta } \right) = {x^2} - \left( { - 1 + \beta } \right)x + \left( { - 1} \right)\beta = {x^2} - \left( {\beta - 1} \right)x - \beta \]
According to the question, the given function satisfies: $f\left( 1 \right) + f\left( 2 \right) = 0$
We can substitute the expression for $f\left( 1 \right)$ and $f\left( 2 \right)$ into the above equation to solve for $\beta $
$ \Rightarrow f\left( 1 \right) + f\left( 2 \right) = 0 \\
\Rightarrow \left( {{1^2} - \left( {\beta - 1} \right) \times 1 - \beta } \right) + \left( {{2^2} - \left( {\beta - 1} \right) \times 2 - \beta } \right) = 0$
Now this equation can be simplified and we can evaluate it further for $\beta $ as:
\[ \Rightarrow \left( {{1^2} - \left( {\beta - 1} \right) \times 1 - \beta } \right) + \left( {{2^2} - \left( {\beta - 1} \right) \times 2 - \beta } \right) = 0 \\
\Rightarrow 1 - \beta + 1 - \beta + 4 - 2\beta + 2 - \beta = 0\]
Let’s shift all the $\beta $ on one side of the equation:
\[ \Rightarrow 1 - \beta + 1 - \beta + 4 - 2\beta + 2 - \beta = 0 \\
\Rightarrow 1 + 1 + 4 + 2 = \beta + \beta + 2\beta + \beta \\
\Rightarrow 5\beta = 8 \\
\Rightarrow \beta = \dfrac{8}{5}\]
Therefore, this gives us the value \[\beta = \dfrac{8}{5}\]
Hence, we have the second root of $f\left( x \right) = 0$ as $\dfrac{8}{5}$

Thus, option (D) is the correct answer.

Note: Be careful while solving the equation $f\left( 1 \right) + f\left( 2 \right) = 0$ . Notice that when we write $f\left( 1 \right)$ we actually just replace the variable $x$ from the function expression and put the value in the parenthesis. An alternative approach for this problem can be to assume the $f\left( x \right) = a{x^2} + bx + c$ and then use the property that the sum of roots can be written as $\dfrac{{ - b}}{a}$ and product of the roots as $\dfrac{c}{a}$ .