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It is possible for a quadratic equation to have no real-number solutions. Solve. t^2+10t+26=0Asked by tom
It is possible for a quadratic equation to have no real-number solutions:Solve : x2 + 5x + 3 = 0
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Answered by
Damon
Sure, that just means that the parabola never crosses the x axis but has a vertex above if it opens up or below if it opens down.
x^2 + 5 x + 3
x = [ -5 +/- sqrt (25 - 12) ] / 2
that has real roots because (b^2-4ac) is positive.
However try
x^2 + 2 x + 3 = 0
x = [ -2 +/- sqrt (4 - 12) ]/2
x = -1 +/- .5 sqrt(-8)
sqrt of -8 = i sqrt 8 = 2i sqrt(2)
x = -1 +/- i sqrt 2
the solutions are complex because they contain i, the sqrt of -1
x^2 + 5 x + 3
x = [ -5 +/- sqrt (25 - 12) ] / 2
that has real roots because (b^2-4ac) is positive.
However try
x^2 + 2 x + 3 = 0
x = [ -2 +/- sqrt (4 - 12) ]/2
x = -1 +/- .5 sqrt(-8)
sqrt of -8 = i sqrt 8 = 2i sqrt(2)
x = -1 +/- i sqrt 2
the solutions are complex because they contain i, the sqrt of -1
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