Question

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

Answer 1

The roots of x^2 + (1 - k).x+k-3 = 0 are real for all real values of k

The equation is given as:

x^2 + (1 - k).x+k-3 = 0

For an equation to have real root, then the following must be true:

$b^2 \ge 4ac$

In x^2 + (1 - k).x+k-3 = 0, we have:

a = 1

b = 1 - k

c = k - 3

So, we have:

$(1 - k)^2 \ge 4 * 1 * (k - 3)$

Evaluate

$1 - 2k + k^2 \ge 4k - 12$

Collect like terms

$k^2 - 2k - 4k + 1 + 12 \ge 0$

$k^2 - 6k + 13 \ge 0$

Using a graphing calculator, the value of k is all real numbers

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

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