Hello people ,

How do i solve these?

1) Find the quadratic equation whose roots are reciprocals of the roots of $x^2-7x+12=0$

2) Find the quadratic equation whose roots are double the roots of $3x^2-4x+7=0$

2. 1) Either by:

a) Solving the equation, taking the reciprocals of the roots (a and b, say), and forming the quadratic equation
$\left({x - \frac 1 a}\right)\left({x - \frac 1 b}\right)$
but that's long-winded.

Alternatively note that for a q.e. $x^2 - px + q$, if the roots are $a$ and $b$, we have:
$a + b = p, ab = q$.

Divide $a + b = p$ by $ab = q$ to get:

$\frac {a+b}{ab} = \frac 1 b + \frac 1 a = \frac p q$

and then note that $\frac 1 {ab} = \frac 1 q$

$x^2 - \frac p q x + \frac 1 q$
and then multiply by $q$ to get the answer.

Plug in your numbers and you're done.

3. Originally Posted by saberteeth
Hello people ,

How do i solve these?

1) Find the quadratic equation whose roots are reciprocals of the roots of $x^2-7x+12=0$

2) Find the quadratic equation whose roots are double the roots of $3x^2-4x+7=0$
for (1) let a and b be the roots of the equation. Then $a+b = 7$ and $ab = 12$. The general procedure is:
The quadratic with reciprocals of the roots will have sum of roots as $\frac1{a} + \frac1{b} = \frac{a+b}{ab} = \frac7{12}$ and product of roots as $\frac1{ab} = \frac1{12}$

You could have also observed 4 and 3 are roots. Thus the required quadratic is obtained by expanding $\left(x - \frac1{3}\right)\left(x - \frac14 \right)$

4. Originally Posted by Matt Westwood
and then note that $\frac 1 {ab} = \frac 1 q$

$x^2 - \frac p q x + \frac 1 {pq}$
I think the equation should read $x^2 - \frac p q x + \frac 1 {q}$

5. Originally Posted by Isomorphism
I think the equation should read $x^2 - \frac p q x + \frac 1 {q}$
Yes it should - I just corrected it. Slip of the brain. Sorry.

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