Find the value of k, for which one root of the quadratic equation kx2 – 14x + 8 = 0 is six times the other.
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Answered by
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Hey mate..
here is your answer
Let roots of the equation are be a and b.
Given a = 6b
now, if a and b are roots then equation is
(x -a)(x -b) =0
=> x² - (a+b)x + ab =0
Now put a = 6b then
x² - (6b +b)x + 6b×b =0
=> x² - 7bx +6b² =0 ...............1
Given equation is
kx² -14x +8 =0
=> kx² - (14/k*x +(8/k) =0 .......................2
Compare equation 1 and 2, we get
7b =14/k
=> b= 14/(7*k)
=> b =2/k
=> b² = 4/k²...............3
and 6b² =8/k
=> 6b² =8/(6k)
=> b² =4/(3k) ...............4
Put value of b² in equation4, we get
4/k² = 4/(3k)
=>1/k = 1/3
=> k =3
So value of k is 3
hope it helps you!!!
please mark my answers as the brainliest answer.
FalakFarheen:
please mark my answers as the brainliest answer.
Answered by
2
Step-by-step explanation:
k=3
thank you for your time
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