the roots of px²-14x+8=0 is 6times the other then p=?
Answers
Hey there !
Solution :
Let the zeros be ' x ' and ' y '
Given that one of the zero is 6 times the other zero. Hence we can write it as:
x = 6y
We know that, in a quadratic equation,
The sum of the zeros = - b / a
Product of zeros = c / a
Here a, b, c refer to the coefficients of x², x and constant respectively.
In the given equation, a = p ; b = - 14 ; c = 8
Sum of zeros = x + y. But we know that x = 6y. Hence we get sum of zeros as y + 6y = 7y .
=> Sum of zeros = 7y
Also Sum of zeros = - b / a
=> Sum of zeros = - ( - 14 ) / p = 14 / p
=> 7y = 14 / p
=> y = 2 / p ----( 1 )
Product of zeros = x ( y )
=> Product = ( 6y ) ( y ) = 6y²
Therefore we know that y = 2 / p. So substituting in the equation we get,
Product = ( 2 / p )² => 4 / p²
Product = c / a
=> Product = 8 / p
So 4 / p² = 8 / p
=> 4p / p² = 8
=> 4 / p = 8
=> p = 4 / 8
=> p = 1 / 2
Hence the value of p is 1 / 2.
Hope my answer helped :-)