Question

Prove that the roots of x² + (1-k)x+k-3=0 are real for all real values of k.​

Answer 1

Step-by-step explanation:To prove that the roots of the equation x² + (1-k)x + k-3 = 0 are real for all real values of k, we need to show that the discriminant of the equation is non-negative for all values of k.

The discriminant of a quadratic equation ax² + bx + c = 0 is given by b² - 4ac. If the discriminant is positive, then the equation has two distinct real roots; if it is zero, then the equation has one real root (a repeated root); and if it is negative, then the equation has no real roots.

So, in this case, the discriminant of the equation is:

(1-k)² - 4(1)(k-3)

= 1 - 2k + k² - 4k + 12

= k² - 6k + 13

We need to show that k² - 6k + 13 ≥ 0 for all real values of k.

To do this, we can complete the square:

k² - 6k + 13

= (k - 3)² + 4

Since the square of any real number is non-negative, we have (k-3)² ≥ 0 for all k, which means that (k-3)² + 4 ≥ 4.

Therefore, k² - 6k + 13 ≥ 4 for all real values of k, which means that the discriminant of the quadratic equation x² + (1-k)x + k-3 = 0 is non-negative for all real values of k. Hence, the roots of the equation are real for all real values of k.

Answer 2

We can prove that the roots of x² + (1-k)x+k-3=0 are real for all real values of k using the discriminant of the quadratic formula. The discriminant is the expression b²-4ac, where a, b, and c are the coefficients of the quadratic equation ax²+bx+c=0.

For the quadratic equation x² + (1-k)x+k-3=0, we have:

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

Using the quadratic formula, we can find the roots of the equation:

x = (-b ± sqrt(b²-4ac))/2a

Substituting in the values of a, b, and c, we get:

x = (-(1-k) ± sqrt((1-k)² - 4(1)(k-3)))/2(1)

Simplifying, we get:

x = (k-1 ± sqrt((k-1)² - 4(k-3)))/2

Expanding the square inside the square root, we get:

x = (k-1 ± sqrt(k² - 2k + 1 - 4k + 12))/2

Simplifying further, we get:

x = (k-1 ± sqrt(k² - 2k + 13))/2

In order for the roots to be real, the expression inside the square root must be non-negative. That is:

k² - 2k + 13 ≥ 0

To find the values of k that satisfy this inequality, we can use the quadratic formula again:

k = (2 ± sqrt(4 - 4(1)(13)))/2

k = 1 ± sqrt(-44)/2

Since the square root of a negative number is not a real number, there are no real values of k that satisfy this inequality. Therefore, the roots of x² + (1-k)x+k-3=0 are real for all real values of k.