Find the value of p, for which one root of the quadratic equation px²-14x+8=0 is 6 time the other.
Answers
Answered by
13
let one root be x
then another root be 6x
then
x + 6x =14/p
7x =14/p
x=2/p
now
x×6x=8/p
6x²=8/p
6×(2/p)²=8/p
4/p²=8/p×6
1/p²= 4/3p×4
1/p²=1/3p
p²=3p
therefore
p=3
then another root be 6x
then
x + 6x =14/p
7x =14/p
x=2/p
now
x×6x=8/p
6x²=8/p
6×(2/p)²=8/p
4/p²=8/p×6
1/p²= 4/3p×4
1/p²=1/3p
p²=3p
therefore
p=3
Himanshukajaria:
answer will be 2
it is given
because alpha + beta = -b/a and b=-14
can't understand
please google the question you can find the correct answer
b=14/p
now it is correct
see the correct process
yes got it before you
thankyou
Answered by
0
let be x value will be 1
p (x)^2-14x+8=0
p (1)^2-14 (1)+8=0
p (1)-14+8 =0
p-14+8 =0
p-6=0
p=6
p (x)^2-14x+8=0
p (1)^2-14 (1)+8=0
p (1)-14+8 =0
p-14+8 =0
p-6=0
p=6
the answer is 3
Similar questions