Question

If one root of the polynomial f(x)=X2 +5x+k is reciprocal of the other.find the value of k and verify it.

Answer 1

Answer:

given that

one root of the quadratic equation is the reciprocal of other

Let Alpha and beta are the roots of the quadratic equation

$x {}^{2} + 5 x + k$

accordinf to the quesion

$\alpha = \frac{1}{ \beta }$

$\alpha \beta = 1$

product of the zeroes

$\alpha \beta = \frac{k}{1}$

putting this value we get

$k = 1$

Answer 2

The value of k is 1.

Step-by-step explanation:

Let the one root of the given polynomial is a.

Then, the other root is reciprocal of a which is $\frac{1}{a}$

Now, the sum of the roots is given by

$a+\frac{1}{a}=-\frac{5}{1}\\\\a+\frac{1}{a}=-5$

And product of roots is given by

$a\cdot\frac{1}{a}=\frac{k}{1}\\\\k=1$

Therefore, the value of k is 1.

Therefore, the polynomial becomes

f(x)=x²+5x+1

Now, find the roots and check if one root is reciprocal of others or not.

$x^2+5x+1=0$

Apply the quadratic formula

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=\frac{-5\pm\sqrt{21}}{2\cdot \:1}\\\\x=\frac{-5+\sqrt{21}}{2},\:x=\frac{-5-\sqrt{21}}{2}$

Now check the roots are reciprocal or not

$\frac{1}{\frac{-5+\sqrt{21}}{2}}\\\\\frac{2}{-5+\sqrt{21}}$

Rationalization: Multiply numerator and denominator with the conjugate

$\frac{2\left(5+\sqrt{21}\right)}{\left(-5+\sqrt{21}\right)\left(5+\sqrt{21}\right)}$

$=\frac{2\left(5+\sqrt{21}\right)}{-4}\\\\=-\frac{5+\sqrt{21}}{2}$

Therefore, both are roots are reciprocal.

Hence, the value of k is 1.

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