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1) Find the quadratic equation whose roots are reciprocal s of the roots of the equation 3x²-20x+17=0.

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## Answers

### This Is a Certified Answer

roots r1 and r2.

r1 + r2 = 20/3 r1 r2 = 17/3

roots of equation needed: 1/r1 and 1/2.

1/r1 + 1/r2 = (r1+r2)/r1r2 = 20/17

product = 1/r1*1/r2 = 3/17

equation x^2 -20/17 x + 3/17 = 0 17 x^2 - 20 x + 3 = 0

2) 3 x^4 - 13 x^2 + 10 = 0

3 x^4 - 3 x^2 - 10 x^2 + 10 = 0

3 x^2(x^2 - 1) - 10 (x^2 -1) = 0

(3 x^2 - 10 ) ( x^2 - 1 ) = 0

x = + root(10/3) - root(10/3) +1 or -1

3) 4 x^2 - 16 x + 15 = 0

4 x^2 - 6 x - 10 x + 15 = 0

2 x ( 2x - 3) - 5 ( 2 x - 3) = 0

(2 x - 5) ( 2 x - 3) = 0

4)

(a-2)^2 + a^2 + (a+2)^2 = 3 a^2 + 8 = 83 => a = 5

5)

5 x - 3 y - 3 = 0

6)

weight man = M weight of child = C

M = 4 C

7)

12 x + 13 y = 29

13 x + 12 y = 21

Multiply 1st equation by 12 and second by 13 and subtract

25 x = 21 * 13 - 29 * 12

find x by simplifying.

1. We have, , we need an equation whose roots ate reciprocal to the roots of f(x), i.e. we need .

Hence required equation is .

2. You have,

3. You have,

Hence,

4. Let the three odd numbers be x-2, x, x+2.

Then

Hence, and numbers are

5. Given equation can be expressed as

Hence, (a, b, c) = (5, -3, -3).

Actually, infinite number of values are possible.

6. Let the weights of man and his child be denoted as y, x.

From the given information,

7. Add the given two equations, you get,

Hence,