Question

The value of ‘k’ for which one of the roots of ${{x}^{2}}-x+3k=0$, is double of one of the roots of ${{x}^{2}}-x+k=0$ is
( a ) 1
( b ) – 2
( c ) 2
( d ) none of these

Answer 1

Hint: In this question, we are given two quadratic equations and one equation is double the other. First, we suppose the roots a, 2a and by putting the roots we find out the value of a, and putting the values of a, we find out the value of k.

Complete step by step Solution:
The given quadratic equations are
${{x}^{2}}-x+3k=0$……………………………………………… (1)
${{x}^{2}}-x+k=0$………………………………………………….(2)
It is given that one of the roots of equation (1) is double of one of the roots of equation (2)
Suppose that x = a be the root of equation (2) then x = 2a will be the root of equation (1)
Since x = a is a root of equation (2)
Then x = a will satisfy the equation (2)
That is ${{a}^{2}}-a+k=0$
Then the value of k = $a-{{a}^{2}}$………………………………………………. (3)
Also x = 2a is a root of equation (1)
Then x = 2a will satisfy the equation (1)
That is ${{(2a)}^{2}}-(2a)+3k=0$
that is $4{{a}^{2}}-2a+3k=0$
we put the value of k from equation (3) in the above equation and get
$4{{a}^{2}}-2a+3(a-{{a}^{2}})=0$
Solving it, we get ${{a}^{2}}+a=0$
Or a (a+1) =0
That is a = 0, -1
If a = 0 then by equation (3)
K = 0 - ${{(0)}^{2}}$
Value of k = 0
If a = -1
Then k = -1 - ${{(-1)}^{2}}$ value of k = -2
Hence the value of k = 0,-2

Therefore, the correct option is (b).

Note: Values which satisfied the equation are called its solution or roots. If x = a is a solution of x, then it must satisfy the given equation otherwise it’s not a root of the equation.