Question

If one zero of polynomial of f(x)=(k^2+4)x^2+13x+14k is reciprocal of other then k is

Answer 1

Question :-

If one zero of polynomial of f(x)=(k^2+4)x^2+13x+14k is reciprocal of other then k is-

Answer:-

→ k = -4 or 10

Step - by - step explanation:-

Used property :-

If the zeros of a polynomial is given then to find the value of any constant in the polynomial we have to used sum of zeros and product of zeros.

Solution :-

Given polynomial is ↓

$\bf{f(x) = ( {k}^{2} + 4) {x}^{2} + 13x + 14k \: }$

Let the zeros of this polynomial are

m and n

Then we know that,

$\red{\bf{product \: of \: zeros \: = \frac{constant}{coefficient \: of \: {x}^{2} } } }$

According to the question,

One zero of the given polynomial is reciprocal of the other zero of the given polynomial.

$\green{ \therefore \: \bf{ m = \frac{1}{n} }}$

$\bf{product \: of \: zeros \: = m \times n \: } \\ \because \: \bf{m = \frac{1}{n} } \\ \\ \therefore \: \\ \bf{ \implies \: \frac{1}{n} \times n = \frac{14 \: k}{( {k}^{2} + 4) } } \\ \\ \implies \: \bf{1 = \frac{14 \: k}{( {k}^{2} + 4) } \: \: } \\ \\ \implies \bf{\: {k}^{2} + 4 = 14 \: k} \\ \\ \bf{ \implies \: {k}^{2} - 14 \: k = - 4} \\ \\ \implies \: \bf{k (k- 14) = - 4}$

• Case (1) if →

$\implies \: \red{\boxed{ \bf{ k = - 4}}}$

• Case (2) if →

$\implies \: \bf{k - 14 = - 4} \\ \\ \implies \: \bf{k = - 4 + 14 }\\ \\ \implies \: \red{ \bf{\boxed{k = 10}}}$

Hence, In the Given polynomial k has two values.

1. k= -4
2. k = 10

Hope it helps you.

Be Brainly : )

Answer 2

$\huge{\boxed{\mathtt{\red{ANSWER}}}}$

roots are reciprocal given, it means , product of roots equal to 1 ..

product of roots = c/a = 14k/(k²+4)

1 = 14k/(k²+4)

k²+4 -14k = 0

k²-14k+4 = 0

K = 7±3√5 (Ans)