Question

can the quadratic polynomial x2+kx+k have equal zeroes for some odd integer k>1

Answer 1

Hey there !

Solution:

For a Quadratic Equation to have equal roots, it must satisfy the condition:

b² - 4ac = 0

Given equation is x² + kx + k = 0

a = 1, b = k, x = k

So Substituting in the equation we get,

=> k² - 4 ( 1 ) ( k ) = 0

=> k² - 4k = 0

=> k ( k - 4 ) = 0

=> k = 0 , k = 4

But in the question, it is given that k is greater than 1.

Hence the value of k is 4 if the equation has common roots.

Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

Hope my answer helped !

Answer 2

Answer:

The quadratic polynomial cannot have equal zeros for any odd integer k > 1   k = 0 or k = 4

Step-by-step explanation:

Concept

• A quadratic polynomial has a second degree monomial as its highest degree monomial. Second-order polynomials are similar to quadratic polynomials. This means that at least one of the variables must be raised to the power of two, with the powers of the other variables being less than or equal to two but greater than -1.

Given

the quadratic polynomial is x² + kx + k

Find

some odd integer

Solution

Given, the quadratic polynomial is x² + kx + k

We have to find whether the zeros of the polynomial are equal for some odd integer k > 1

Assuming the zeros of the polynomial are equal,

The value of the discriminant .

Discriminant = b² - 4ac

b² - 4ac = 0

b² = 4ac

Here, a = 1, b = k and c = k

(k)² - 4(1)(k) = 0

k² - 4k = 0

k(k - 4) = 0

So, k = 0 or k = 4.

The quadratic polynomial will have equal zeros at k = 0 and k = 4

Therefore, the quadratic polynomial cannot have equal zeros for any odd integer k > 1

#SPJ2