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 [UPSEAT 2001]
A) 1 B) - 2 C) 2 D) None of these

Answer 1

Hi ,

Let is m is the root of x² - x + 3k = 0

m² - m + 3k = 0 ---( 1 )

according to the problem given ,

2m is the one root of x² - x + k = 0

( 2m )² - 2m + k = 0

4m² - 2m + k = 0 ----- ( 2 )

( 1 ) = ( 2 )

m² - m + 3k = 4m² - 2m + k

3k - k = 4m² - 2m - m² + m

2k = 3m² - m

k = ( 3m² - m )/2 --( 3 )

put k value in equation ( 1 ) , we get

m² - m + 3( 3m² - m )/2 = 0

[2m² - 2m + 3(3m² - m) ]/2 = 0

2m² -2m + 9m² - 3m = 0

11m² -5m = 0

m ( 11m - 5 ) = 0

m = 0 or 11m - 5= 0

m = 5/11

Option ( D ) is correct.

I hope this helps you.

: )

Answer 2

Answer: b) - 2

Step-by-step explanation: Let one root of x2 - x + 3k = 0 is 2m

Sub 2m in the equation

4m2 - 2m + 3k = 0 - - - eq (1)

Now, let one root of equation x2 - x + k = 0 is m

Sub m in the equation

m2 - m + k = 0 - - - eq (2)

Multiply the eq (2) with 4

4m2 - 4m + 4k = 0 - - - eq (3)

Now, (1) = (3)

4m2 - 2m + 3k = 4m2 - 4m +4k

2m - k = 0

m = k/2

Sub m value in eq (2)

(k/2)2 - k/2 + k = 0

(k2 - 2k + 4k) /2 = 0

k2 + 2k = 0

k(k+2) = 0

k = - 2