Question

If one zeroes of quadratic polynomial. (K -1 )X² + Kx + 1 is 1 than K is

(1 Point)

Zero

1

-1

2

​

Answer 1

Given us a quadratic polynomial,

• p(x) = (k - 1)x² + kx + 1

One of its two zeroes is 1, so we have to find the value of k.

we know, a zero of a polynomial is a value that satisfies that polynomial. In other words, it makes the polynomial zero.

So, Substituting x = 1 , p(x) will be 0. we have

⇒ p(1) = 0

⇒ (k - 1)(1)² + k(1) + 1 = 0

⇒ (k - 1) + k + 1 = 0

⇒ k - 1 + k + 1 = 0

⇒ 2k = 0

⇒ k = 0

Verification :-

Put k = 0 in p(x) and then substitute x = 1

⇒ p(x) = (0 - 1)x² + (0)x + 1

⇒ p(x) = -x² + 1

Now, Substituting x = 1 , we must get 0

⇒ p(1) = -(1)² + 1

⇒ 0 = -1 + 1

⇒ 0 = 0

Hence, Verified.

Some Information :-

☞ A quadratic polynomial is a polynomial of degree 2. where degree is the highest power of the variable of that polynomial.

☞ The degree of the polynomials also tells us the number of solutions of that polynomial. If a polynomial has degree n then there will be n number of solutions of that polynomial.

Answer 2

Answer:

▶ K = 0

Step-by-step explanation:

Given that,

▶ Quadratic polynomial p(x) = (k - 1)x² + kx + 1.

Substituting x = 1 : We get,

▶p(x = 1) = 0

▶(k - 1)(1) + k(1) + 1 = 0

▶k - 1 + k + 1 = 0

▶k + k - 1 + 1 = 0

▶2k = 0

▶k = 0 × 2

▶k = 0

Hence,

• The value of k is 0.