Question

If both the roots of ${k}(6{x^2} + 3) + rx + 2{x^2} - 1 = 0$ and $6k(2{x^2} + 1) + px + 4{x^2} - 2 = 0$ are common, $2r - p$ is equal to
A. $ - 1$
B. $0$
C. $1$
D. $2$

Answer 1

Hint: We are given two quadratic equations which possess roots and both the roots are common. To get the result, write the equations in the decreasing power of variable $x$ . Then, compare the coefficients of both the equations and substitute in the condition for common roots.
Formula Used: Suppose the two quadratic equations are: ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ , then the condition for common roots will be-
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ .

Complete step-by-step solution:
We have two quadratic equations such as
${k}(6{x^2} + 3) + rx + 2{x^2} - 1 = 0$
And
$6k(2{x^2} + 1) + px + 4{x^2} - 2 = 0$
We can also write both the equations as
$(6k + 2){x^2} + rx + 3{k} - 1 = 0$
Similarly
$2(6k + 2){x^2} + px + 2(3k - 1) = 0$
Now, if we compare these equations with the general quadratic equations ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ , we get the values of the coefficients
${a_1} = 6k + 2$ , ${b_1} = r$ , ${c_1} = 3k - 1$
${a_2} = 12k + 4$ , ${b_2} = p$ , ${c_2} = 6k - 2$
Substituting these values in the condition for common roots i.e.,
$\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{b_2}}}{{{b_1}}} = \dfrac{{{c_2}}}{{{c_1}}}$ , we get
$\dfrac{{12k + 4}}{{6k + 2}} = \dfrac{p}{r} = \dfrac{{6k - 2}}{{3k - 1}} = 2$
Equalizing the second and fourth parts of this equation, we get
$\dfrac{p}{r} = 2$ , which on solving gives
$2r - p = 0$ .
Hence, the correct option is B.

The kind of solution or roots depends on the discriminant's value. $D = {b^2} - 4ac$ can be used to calculate discriminant. If the value $D > 0$ , the roots will be real and distinct. If $D = 0$ , then the roots will be real and equal. Lastly, if the value $D < 0$ , then the roots will be imaginary and pairwise.