# The graph of a quadratic function has x-intercepts -2 and 7/2, how do you write a quadratic equation that has these roots?

##### 1 Answer

Find f(x) = ax^2 + bx + c = 0 knowing the 2 real roots: x1 = -2 and x2 = 7/2.

Given 2 real roots c1/a1 and c2/a2 of a quadratic equation ax^2 + bx + c = 0, there are 3 relations:

a1*a2 = a
c1*c2 = c

a1

*c2 + a2c1 = -b (Diagonal Sum).*

In this example, the 2 real roots are: c1/a1 = -2/1 and c2/a2 = 7/2.

a = 12 = 2

In this example, the 2 real roots are: c1/a1 = -2/1 and c2/a2 = 7/2.

a = 1

c = -2

*7 = -14*

-b = a1c2 + a2c1 = -22 + 1

-b = a1c2 + a2c1 = -2

*7 = -4 + 7 = 3.*

The quadratic equation is:

Answer: 2x^2 - 3x - 14 = 0 (1)

Check: Find the 2 real roots of (1) by the new AC Method.

Converted equation: x^2 - 3x - 28 = 0 (2). Solve equation (2). Roots have different signs. Compose factor pairs of ac = -28. Proceed: (-1, 28)(-2, 14)(-4, 7). This last sum is (-4 + 7 = 3 = -b). Then its 2 real roots are: y1 = -4 and y2 = 7. Back to original equation (1), the 2 real roots are: x1 = y1/a = -4/2 = -2 and x2 = y2/a = 7/2. Correct.

The quadratic equation is:

Answer: 2x^2 - 3x - 14 = 0 (1)

Check: Find the 2 real roots of (1) by the new AC Method.

Converted equation: x^2 - 3x - 28 = 0 (2). Solve equation (2). Roots have different signs. Compose factor pairs of a