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

2) solve the following quadratic equation by factorisation method. 3x⁴-13x²+10=0. 3) solve 4x²-16x+15=0 by completing square method.
4) The sum of the squares of the three consecutive old numbers is 83,find the numbers.

And now this of Linear Equations in two Variables.

5) Express linear equation 3y=5x-3 in the form ax+by+c=0 and write the value of a,b & c.
6) The weight of man is a four times the weight of his child.Write an equation in two variables for this situation.
7) If 12x+13y=29 & 13x+12y=21 find (x+y).

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2015-01-04T13:35:34+05:30

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1)  3 x^2  - 20 x + 17 = 0 
     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.

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  • Brainly User
2015-01-04T14:44:18+05:30

1. We have, f(x)=3x^{2}-20x+17=0 , we need an equation whose roots ate reciprocal to the roots of f(x), i.e. we need f(\frac{1}{x})=3(\frac{1}{x})^{2}-20(\frac{1}{x})+17.
Hence required equation is 17x^{2}-20x+3=0.

2. You have,
3x^{4}-13x^{2}+10=0
3x^{2}(x^{2}-1)-10(x^{2}-1)=0
x=-\sqrt{\frac{10}{3}}, +\sqrt{\frac{10}{3}}, -1, +1.

3. You have,
(2x)^{2}-2(2x)(4)+(4^{2})-1=0
(2x-4)^{2}=1
Hence, x=\frac{5}{2}, \frac{3}{2}.

4. Let the three odd numbers be x-2, x, x+2.
Then (x-2)^{2}+(x^{2})+(x+2)^{2}=83
3x^{2}=75
Hence, x=5 and numbers are 3, 5, 7.

5. Given equation can be expressed as 5x-3y-3=0
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, y=4x

7. Add the given two equations, you get,
25(x+y)=50
Hence, x+y=2

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