Question

if alpha,beta are the zeroes of polynomial kx2+5x+2 such that 1/alpha2+1/beta2 = 17/4 , find k

Answer 1

Step-by-step explanation:

above is the answer of the question...

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Answer 2

The correct answer is 2.

Given: The equation = $kx^{2} +5x+2=0$.

$\frac{1}{a^{2} } +\frac{1}{b^{2} }=\frac{17}{4}$.

To Find: The value of k.

Solution:

$\frac{1}{a^{2} } +\frac{1}{b^{2} }=\frac{17}{4}$

$\frac{a^{2} +b^{2} }{a^{2} b^{2} }=\frac{17}{4}$

$\frac{(a+b)^{2}-2ab }{(ab)^{2} }=\frac{17}{4}$   (equation 1)

Sum of roots  = a + b

= $\frac{-(coefficient of x)}{coefficient of x^{2} }$

a + b = $\frac{-5}{k}$

Product of roots = ab

= $\frac{constant}{coefficient of x^{2} }$

ab = $\frac{2}{k}$

Put these values in equation 1.

$\frac{(\frac{-5}{k} )^{2}-2(\frac{2}{k} ) }{(\frac{2}{k} )^{2} } =\frac{17}{4}$

$\frac{\frac{25}{k^{2} }-\frac{4}{k} }{\frac{4}{k^{2} } } =\frac{17}{4}$

$\frac{25-4k }{4} =\frac{17}{4}$

25 - 4k = 17

4k = 8

k = 2

Hence, the value of k is 2.

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