Question

38.If a and b are the roots of the equation x^2-3x-2 = 0 ,form an equation whose roots are 1/(2a+b) and 1/(2b + a)​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given:

If a and b are the roots of the equation x^2-3x-2 = 0 ,form an equation whose roots are 1/(2a+b) and 1/(2b + a)

To find:equation whose roots are 1/(2a+b) and 1/(2b + a)

Solution:

In order to form the equation whose roots are 1/(2a+b) and 1/(2b + a)

First find out the value of a and b from given equation

Find roots of x²-3x-2=0 by quadratic formula

$x_{1,2}= \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ here \\ a = 1 \\ b = - 3 \\ c = - 2 \\ \\ x_{1,2} = \frac{ - ( - 3)± \sqrt{ {( - 3)}^{2} - 4(1)( - 2)} }{2 \times 1} \\ \\ x_{1,2} = \frac{3 ± \sqrt{9 + 8} }{2} \\ \\ x_{1,2} = \frac{3 ± \sqrt{17} }{2}$

Since root of equations are a and b

$a = \frac{3 + \sqrt{17} }{2 } \\ \\ b = \frac{3 - \sqrt{17} }{2 }$

Find the value of sum of

$= \frac{1}{2a + b} + \frac{1}{2b + a}$

Solve

$\frac{1}{2a + b} = \frac{1}{2 \times ( \frac{3 + \sqrt{17} }{2}) + ( \frac{3 - \sqrt{17} }{2})} \\ \\ \frac{1}{2a + b} = \frac{1}{3 + \sqrt{17} + \frac{3}{2} - \frac{ \sqrt{17} }{2}} \\ \\ \frac{1}{2a + b} = \frac{2}{9 + \sqrt{17} }$

by the same way

$\frac{1}{2b + a} = \frac{2}{9 - \sqrt{17} }$

$\frac{1}{2a + b} + \frac{1}{2b + a} = \frac{2}{9 + \sqrt{17} } + \frac{2}{9 - \sqrt{17} } \\ \\ = \frac{2(9 - \sqrt{17}) + 2(9 + \sqrt{17} )}{(9 + \sqrt{17})(9 - \sqrt{17} ) } \\ \\ = \frac{18 - 2 \sqrt{17} + 18 + 2 \sqrt{17} }{81 - 17} \\ \\ = \frac{36}{64} \\ \\ \frac{1}{2a + b} + \frac{1}{2b + a} = \frac{9}{16}$

Product of zeroes

$\frac{1}{2a + b} \times \frac{1}{2b + a} = \frac{2}{9 + \sqrt{17} } \times \frac{2}{9 - \sqrt{17} } \\ \\ = \frac{4}{64} \\ \\ \frac{1}{2a + b} \times \frac{1}{2b + a} = \frac{1}{16}$

If sum of zeroes and product of zeros are known then equation of quadratic polynomial is given by

${x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0 \\ \\ {x}^{2} - \frac{9}{16} x + \frac{1}{16} = 0 \\ \\ 16 {x}^{2} - 9x + 1 = 0$

Thus,

The equation is

$16 {x}^{2} - 9x + 1 = 0$

Hope it helps you.

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1)find the quadratic equation whose root are the reciprocals of the root of x^2+4x-10=0

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2)Find the quadratic equation whose roots are the reciprocals of the roots of

x² +4x - 10 =0.

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