Find the value of p, for which one root of the quadratic equation px2 -14x + 8 = 0 is 6 times the othe
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Hi ,
Let r , 6r are two roots of
px² - 14x + 8 = 0 ,
Compare above equation with
ax² + bx + c = 0
a = p , b = -14 , c = 8
Sum of the roots = - b/a
r + 6r = - ( - 14 ) / p
7r = 14/p
r = 2/p -----( 1 )
Product of the roots = c/a
r × 6r = 8/p
3r² = 4/p
3 × ( 2/p )² = 4 /p [ from ( 1 ) ]
3 × 4/p² = 4/p
3/p = 1
3 = p
p = 3
I hope this helps you.
:)
Let r , 6r are two roots of
px² - 14x + 8 = 0 ,
Compare above equation with
ax² + bx + c = 0
a = p , b = -14 , c = 8
Sum of the roots = - b/a
r + 6r = - ( - 14 ) / p
7r = 14/p
r = 2/p -----( 1 )
Product of the roots = c/a
r × 6r = 8/p
3r² = 4/p
3 × ( 2/p )² = 4 /p [ from ( 1 ) ]
3 × 4/p² = 4/p
3/p = 1
3 = p
p = 3
I hope this helps you.
:)
Similar questions
let a and b be the roots of the equation
b=6a
sum of roots = -b/a
product of roots = c/a
here a=p , b=-14 , c=8
a+b=14/p
ab=8/p
a+6a=14/p
7a=14/p
a=2/p
a*6a=8/p
6a^2=8/p
3a^2=4/p
3*(2/p)^2=4/p
3*4/p^2=4/p
p=3