# For what values of k does the function f(x) = x^2 + kx - 5 have two real roots?

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### 2 Answers

If the quadratic function f(x) = x^2 +kx -5 has two real roots, the term b^2 - 4*a*c should be greater than 0

Here b = k , a = 1 and c = -5

=> k^2 + 4*5*1 > 0

=> k^2 > -20

But k^2 is always a positive number and hence greater than -20.

**Therefore k can take on any value for the function f(x) = x^2 +kx -5 to have 2 real roots.**

Given the quadratic equation f(x) = x^2 + kx -5

We need to find the values of k where the function has 2 real roots.

We know that the function has 2 real roots when delta > 0

delta = b^2 - 4ac > 0

==> a = 1 b= k c = -5

==> k^2 - 4*1*-5 > 0

==> k^2 + 20> 0

But we know that k^2 is always > 0

also 20 > 0

Then k^2 + 20 > 0 for all real numbers.

**==> K belongs to R where R is a real number.**

**==> K = (-inf, inf)**