The graph of a quadratic function has x-intercepts -2 and 72, how do you write a quadratic equation that has these roots?

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ocretz56 · Accepted Answer

Step 1  Given 2 real roots c1a1 and c2a2 of a quadratic equation ax2+bx+c=0, there are 3 relations:  a1a2=a  c1c2=c  a1c2+a2c1=−b (Diagonal Sum)  In this example, the 2 real roots are: c1a1=−21 and c2a2=72.  a=12=2  c=−27=−14  −b=a1c2+a2c1=−22+17=−4+7=3.  The quadratic equation is:  Answer: 2x2−3x−14=0 (1)  Check: Find the 2 real roots of (1) by the new AC Method.  Converted equation: x2−3x−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=y1a=−42=−2  and  x2=y2a=72. Correct.

The graph of a quadratic function has x-intercepts -2 and

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