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Find a quadratic equation with the two given numbers as solutions: 2 and -5
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x^2 +3x -10
You can go backwords: (x-2)(x+5) = x^2 +3x -10 and to check the answer, you can use the quadratic equation: \[( -b \pm \sqrt{b^2 - 4ac} ) /2z\] Which gives us \[(-3 \pm \sqrt{(-3)^2 - (4)(1)(-10)} ) / 2a\]
\[-6 + 3i \sqrt{5}\] and \[-6 - 3i \sqrt{5}\]
droid, the solution isn't complex....
different question sorry
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Which results in (-3 +- sqrt { 9 +40 } ) / 2 So we have (-3 +- sqrt {49 }) / 2 which gives us (-3 +- 7) / 2 And in case 1 we have: x = (-3 + 7)/ 2 x = 4 / 2 = 2 And in case 2 we have: x = (-3 - 7) / 2 x = -10 / 2 = -5 And so we see that our equation holds.
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