X ^ 2 + 4x-5 find the value of x at which the function takes positive values and negative ones.

Equate the expression to zero and find the roots of the quadratic equation.

x ^ 2 + 4x – 5 = 0.

D = 4 * 4 + 20 = 36 = 6 ^ 2.

x1 = (- 4 + 6) / 2 = 1;

x2 = (- 4 – 6) / 2 = – 5.

Let’s use the formula and represent the function as factors:

ax ^ 2 + bx + c = a (x – x1) (x – x2).

x ^ 2 + 4x – 5 = (x – 1) (x + 5) find the intervals on which the function takes positive values:

(x – 1) (x + 5)> 0;

x – 1> 0;

x + 5> 0;

also

x – 1 <0;

x + 5 <0;

Answer: the function is positive for all x Є] – ∞: -5 [U] 1; + ∞ [and is negative on the interval] – 5; 1[.