Question

If one of the zero of the polynomial 4x2-17x +k is reciprocal of other find k

Answer 1

Let α be the first root of the polynomial 4x²-17x +k

Then according to question other root is 1/α

Now, product of roots=c/a

α× 1/α= k/4

k=4

Answer 2

Recall the concept of zeros of a polynomial

The zero of a polynomial is the value that makes the value of the polynomial equal to zero.

For a second-degree polynomial, there will be two zeros.

Graphically, the zeros are values at which the curve touches the coordinate axis.

Given:

A polynomial $4x^2-17x+k$

The one root is reciprocal of the other.

Let one root be $a$ and the other root be $\frac{1}{a}$

Since roots make the polynomial value equal to zero,

$\implies 4(a)^2-17(a)+k=0$

$\implies 4a^2-17a+k=0$       ................(1)

Also,

$\implies 4(\frac{1}{a} )^2-17(\frac{1}{a} )+k=0$

$\implies \frac{4}{a^2} -\frac{17}{a} +k=0$         ..........(2)

Subtracting (1) and (2)

$\implies 4a^2-17a+k-\frac{4}{a^2} +\frac{17}{a} -k=0$

$\implies 4a^2-\frac{4}{a^2}-17a +\frac{17}{a} =0$

$\implies 4(a^2-\frac{1}{a^2})-17(a -\frac{1}{a}) =0$

$\implies 4(a^2-\frac{1}{a^2})=17(a -\frac{1}{a})$

Using the identity $a^2-b^2=(a+b)(a-b)$

$\implies 4(a-\frac{1}{a})(a+\frac{1}{a} )=17(a -\frac{1}{a})$

$\implies 4(a+\frac{1}{a} )=17$

$\implies 4(\frac{a^2+1}{a} )=17$

$\implies 4a^2+4=17a$

$\implies 4a^2-17a+4=0$

Using splitting the mid-term,

$\implies 4a^2-16a-a+4=0$

$\implies 4a(a-4)-1(a-4)=0$

$\implies (4a-1)(a-4)=0$

$\implies a=4,\frac{1}{4}$

Substituting the value any of $a$ in equation (1)

$\implies 4(4)^2-17(4)+k=0$

$\implies 4(16)-17(4)+k=0$

$\implies 4(16-17)+k=0$

$\implies -4+k=0$

$\implies k=4$

Hence the value of $k=4$