Question

Prove that the roots of x^2 +(1-k)x+k-3=0 are real for all real values of k

Answer 1

Answer:

a = 1 b = (1-k) c = (k-3)

Step-by-step explanation:

Using Quadratic formula of discriminant

D = B^2- 4 A C

Answer 2

The roots are not real.

Step-by-step explanation:

To prove : The roots of $x^2 +(1-k)x+k-3=0$ are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. $b^2-4ac\geq 0$

The quadratic equation $x^2 +(1-k)x+k-3=0$

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

$D=(1-k)^2-4(1)(k-3)$

$D=1+k^2-2k-4k+12$

$D=k^2-6k+13$

$D=(k-(3+2i))(k+(3+2i))$

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

#Learn more

If x²+kx+k=0, find the value of k, If the roots are real & equal.​

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