Question

If $a,b$are roots of the equation$2{x^2} - 35x + 2 = 0$, then the value of${\left( {2a - 35} \right)^3}{\left( {2b - 35} \right)^3}$.
A) 10
B) 100
C) 64
D) 1

Answer 1

This question can be solved by firstly finding the product of the roots of the equation which equals$\dfrac{c}{a}$, and then by putting the roots $a,b$into the equation one by one and thereby simplifying the further equations, we can easily find the values for $\left( {2a - 35} \right)$and $\left( {2b - 35} \right)$thus, finally ${\left( {2a - 35} \right)^3}{\left( {2b - 35} \right)^3}$can be computed.

Formula used: The formula used here is firstly for the product of roots
i.e. the product of roots is given by
$ab = \dfrac{c}{a}$
Then the quadratic equation given is simplified by rearranging the terms and taking the terms common.
And then, by putting both the roots into the equation one by one, we can find the values of $\left( {2a - 35} \right)$and$\left( {2b - 35} \right)$, the question is solved.

Complete step-by-step solution:
First of all let us write the equation given in the question,
$2{x^2} - 35x + 2 = 0$
As it is given that $a$and $b$are the roots of the equation, then the product of roots is given by the equation,
$ab = \dfrac{c}{a} = \dfrac{2}{2} = 1$
Now, putting $a$as the root in the equation,
$2{a^2} - 35a + 2 = 0$
Now, simplifying this equation further
$\Rightarrow 2{a^2} - 35a + 2 = 0 \\\ \Rightarrow 2{a^2} - 35a = - 2 \\\ \Rightarrow a\left( {2a - 35} \right) = - 2 \\\ \Rightarrow 2a - 35 = \dfrac{{ - 2}}{a} \\\$
We get the value of the first factor which we needed to find.
Now putting $b$ as the root of the equation given
$\Rightarrow 2{b^2} - 35b + 2 = 0 \\\ \Rightarrow 2{b^2} - 35b = - 2 \\\ \Rightarrow b\left( {2b - 35} \right) = - 2 \\\ \Rightarrow 2b - 35 = \dfrac{{ - 2}}{b} \\\$
Here, we got the value of the second factor.
Now writing the equation, the value of which we need to calculate.
${\left( {2a - 35} \right)^3}{\left( {2b - 35} \right)^3}$
Using the above obtained values in this equation, we get
$\Rightarrow {\left( {2a - 35} \right)^3}{\left( {2b - 35} \right)^3} \\\ \Rightarrow {\left( {\dfrac{{ - 2}}{a}} \right)^3}{\left( {\dfrac{{ - 2}}{b}} \right)^3} \\\ \Rightarrow \left( {\dfrac{{ - 8}}{{{a^3}}}} \right)\left( {\dfrac{{ - 8}}{{{b^3}}}} \right) \\\ \Rightarrow \dfrac{{64}}{{{a^3}{b^3}}} \\\$
Now, as we estimated the value of $ab$above, using the value $ab = 1 \Rightarrow {a^3}{b^3} = 1$in the above equation, we get
$\Rightarrow \dfrac{{64}}{1} = 64$

The factor ${\left( {2a - 35} \right)^3}{\left( {2b - 35} \right)^3}$equals to the value $64$

Note: It is important to note that the step wherein we are taking the two roots $a$and $b$separately in the quadratic equation and finding the values of $\left( {2a - 35} \right)$and $\left( {2b - 35} \right)$ separately first and then putting in the main equation. In this the mistake can be made if during finding the value of either$\left( {2a - 35} \right)$or $\left( {2b - 35} \right)$, but the right values give the perfect result.