Question

prove that the zeros of the quadratic polynomial X square + kx + K, K is not equal to zero can't be both positive

Answer 1

Answer:

Step-by-step explanation:

P(x)=x^2+Kx+K

Comparing :ax^2+bx+c

D=b^2-4ac

=(K)^2-4(1)(K)

=K^2-4K

By quadratic formula:

Roots= -(b)+√D. and. =-(b)-✓D

______. ______

2a. 2a

= -K+√K^2-4K. =-K -√K^2-4K

™ __________. _________

2*1. 2*1

=-K+√K(K-4). =-K-√K(K-4)

________. ________

2. 2

Here √K(K-4)>0

In quadratic polynomial ax2 + bx + c

If a > 0, b> 0, c> 0 or a< 0, b< 0,c< 0,

then the polynomial has always all negative zeroes.

and if a > 0, c < 0 or a < 0, c > 0, then the polynomial has always zeroes of opposite sign

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.

Answer 2

Let p(x) = x2 + kx + k, k≠0

On comparing p(x) with ax2 + bx + c, we get

Here, we see that

k(k − 4)> 0

⇒ k ∈ (-∞, 0) u (4, ∞)

Now, we know that

In quadratic polynomial ax2 + bx + c

If a > 0, b> 0, c> 0 or a< 0, b< 0,c< 0,

then the polynomial has always all negative zeroes.

and if a > 0, c < 0 or a < 0, c > 0, then the polynomial has always zeroes of opposite sign

Case I If k∈ (-∞, 0) i.e., k<0

⇒ a = 1>0, b,c = k<0

So, both zeroes are of opposite sign.

Case II If k∈ (4, ∞)i.e., k≥4

⇒ a = 1> 0, b,c>4

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.