Hello people ,
How do i solve these?
1) Find the quadratic equation whose roots are reciprocals of the roots of
2) Find the quadratic equation whose roots are double the roots of
1) Either by:
a) Solving the equation, taking the reciprocals of the roots (a and b, say), and forming the quadratic equation
but that's long-winded.
Alternatively note that for a q.e. , if the roots are and , we have:
.
Divide by to get:
and then note that
So your equation is then:
and then multiply by to get the answer.
Plug in your numbers and you're done.
for (1) let a and b be the roots of the equation. Then and . The general procedure is:
The quadratic with reciprocals of the roots will have sum of roots as and product of roots as
You could have also observed 4 and 3 are roots. Thus the required quadratic is obtained by expanding
1) Find the quadratic equation whose roots are reciprocals of the roots of
we find the roots by factorisation
12 = (-4) *(-3) -7 = -4 - 3
we split the middle term
x^2 -7x +12 = 0
x^2 -4x -3x +12 = 0
x(x-4) -3 (x-4) = 0
(x-3) (x-4) = 0
so the roots are x-3 = 0 gives x =3
and x-4 = 0 gives x =4
Roots are 3,4
Reciprocal of the roots are 1/3,1/4
Quadratic equation: x^2 - (sum of roots) x + (product of roots) = 0
Sum of roots =1/3 +1/4 = 4/12 +3/12 = 7/12
product of roots = 1/3 * 1/4 = 1/12
so the quadratic equation is x^2 - 7/12 x + 1/12 = 0
12 x^2 -7x + 1 = 0
2) Find the quadratic equation whose roots are double the roots of
3x^2 -4x +7 = 0
a = 3 , b =-4 and c =7
sum of roots = - b / a
product of roots = c/a
sum of roots = 4/3
let one root is alpha and the other root is beta
that is alpha + beta = 4/3
alpha * beta = 7/3
for the quadratic equation,
now the roots are doubled means 2* alpha and 2 * beta
are the roots
sum of roots = (2 * alpha) + (2 * beta) = 2(alpha + beta) = 2 * 4/3 = 8/3
product of roots = (2*alpha)(2*beta) = 4 (alpha * beta) = 4 * 7/3 = 28/3
so the quadratic equation is x^2 - 8/3 x + 28/3 =0
3 x^2 - 8x + 28 = 0