Question

a and 1/a are zeroes of polynomial 4x2-2x+k+7. Find the value of k

Answer 1

Solution :

According to the data provided to us the  polynomial where  a and 1/a are it zeroes 4x^2 - 2x + k + 7 and asked to find the value of k

We Know that ,

Product of the zeroes = $\frac{c}{a}$

In the quadratic equation given to us ,

" a "Stands for coefficient of x^2 of the quadratic equation

"b" Stands for coefficient of x of the quadratic equation

"c" Stands for constant term of the quadratic equation

Applying the Product of zeroes in order to get subjection ends in finding the value of k

⇒ a × $\frac{1}{a}$ = $\frac{K +7}{4}$

⇒ $\frac{K + 7}{4} = 1$

⇒ K + 7 = 4

⇒ K = 4 - 7

⇒ K = -3

Therefore for the given quadratic polynomial where two of it's zeroes are a and 1/a the value of k of that is equal to -3

Answer 2

⇨GIVEN:-

$\tt ↠ \red{2x = 3 + \sqrt{7} }$

⇨FIND:-

$\tt ↠ \blue{ {4x}^{2} + \frac{1}{ {x}^{2} } }$

⇨SOLUTION:-

$\tt⤳ \green{2x = 3 + \sqrt{7} }.....(i)$

$\tt➾ \frac{1}{x} = \frac{2}{3 + \sqrt{7} } \\ \tt \gray{❅ rationalise \: the \: denominator} \\ \tt ➾ \frac{2}{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} } \\۞\tt by \: using \: identity \: (a + b)(a - b) = {a}^{2} - {b}^{2} \: we \: have\\ \tt \frac{2(3 - \sqrt{7} )}{ {3}^{2} - ( \sqrt{7} {)}^{2} } = \frac{2(3 - \sqrt{7} )}{ 9 - 7 } \\ \tt➾ \frac{ \cancel2(3 - \sqrt{7} )}{ \cancel2} \\ \tt ➾ \pink{ \frac{1}{x} = 3 - \sqrt{7}} .....(ii)$

adding eq(i) and (ii) we have,

$\tt⪼2x + \frac{1}{x} = 3 \cancel{+ \sqrt{7} } + 3 \cancel{- \sqrt{7} }$

$\tt⪼2x + \frac{1}{x} = 3 + 3 \\ \tt⪼2x + \frac{1}{x} = 6$

now, squaring both sides we have,

$\tt \implies {(2x + \frac{1}{x}) }^{2} = {6}^{2}$

$\tt ✼by \: using {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: we \: have$

$\tt \implies {(2x)}^{2} + \frac{(1 {)}^{2} }{ {x}^{2} } + 2 \times 2 \cancel x \times \frac{1}{ \cancel x } = 36$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } + 4 = 36$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 4$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } = 32$

$\tt Hence, \huge\boxed{ \bold{ 4 {x}^{2} + \frac{1}{ {x}^{2} } = 32}}$