Question

what is the answer of this question?​

Answer 1

Given :

Polynomial -

• $\tt\:f(x)~=~(k^{2}+4)x^{2}+13x+4k$

➷ It is also given that one zero of the given polynomial is reciprocal of the other.

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To find :

• Value of k = ?

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Concept :

For this we first need to assume the zeroes of the given polynomial. Then after , using the formula which follows :

❒ $\tt\:Product~of~the~zeroes~oR~roots~=~\frac{c}{a}$

we could calculate the required value of k.

Hope I sound clear let's do it :D !~

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Assumption :

Let one zero of the polynomial be $\tt\alpha$

Then according to the question :

• Other zero = $\tt\:\frac{1}{\alpha}$

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Solution :

Given polynomial :

$\tt\:f(x)~=~(k^{2}+4)x^{2}+13x+4k$

Here :

• a = $\tt\:k^{2}+4$

• b = $\tt\:13$

• c = $\tt\:4k$

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So now applying the formula :

❒ $\tt\:Product~of~the~zeroes~oR~roots~=~\frac{c}{a}$

Substituting the values

$\tt\implies\:\alpha \times \frac{1}{\alpha}~=~\frac{4k}{k^{2}+4}$

$\tt\implies\:\frac{\alpha}{\alpha}~=~\frac{4k}{k^{2}+4}$

$\tt\implies\:\cancel{\frac{\alpha}{\alpha}}~=~\frac{4k}{k^{2}+4}$

$\tt\implies\:1~=~\frac{4k}{k^{2}+4}$

Cross multiplying

$\tt\implies\:k^{2}+4~=~4k$

Transposing +4k from RHS to LHS it becomes -4k

$\tt\implies\:k^{2}-4k+4~=~0$

Now we have arrived to the quadratic equation , so solving this equation by middle term breaking

$\tt\implies\:k^{2}-(2+2)k+4~=~0$

$\tt\implies\:k^{2}-2k-2k+4~=~0$

Taking k common from first two terms and 2 from second two terms

$\tt\implies\:k(k-2)-2(k-2)~=~0$

Taking ( k-2 ) common from whole expression

$\tt\implies\:(k-2)(k-2)~=~0$

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$\tt\therefore\:k-2=0$

$\small{\underline{\boxed{\mathrm\purple{\dashrightarrow\:k~=~2}}}}$ ★

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Answer 2

Answer

Given :

Polynomial -

•

➷ It is also given that one zero of the given polynomial is reciprocal of the other.

‎

‎

To find :

Value of k = ?

‎

ANSWER

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$\tt\:f(x)~=~(k^{2}+4)x^{2}+13x+4k$