SOLUTION: For what values of k does 2x^2 - 5x - k = 0 have no real root. Thanks so much in advance:)

Question 825644: For what values of k does 2x^2 - 5x - k = 0 have no real root.
Thanks so much in advance:)
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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2x^2 - 5x - k = 0
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first, analyze this:
y(x) = 2x^2 - 5x = 0
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the above quadratic equation is in standard form, with a=2, b=-5, and c=0
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
2 -5 0
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a minimum at: ( x= 1.25, y= -3.125 )
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now, analyze this:
2x^2 - 5x - k = 0
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for (k < -3.125) the equation (2x^2 - 5x - k = 0) will have complex roots (i.e. non-real roots.)
NOTE:
-3.125 is the y-coordinate of the quadratic vertex above
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check:
k < -3.125
arbitrarily set k = -3.1250000001
2x^2 - 5x + 3.1250000001 = 0
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the above quadratic equation is in standard form, with a=2, b=-5, and c=3.1250000001
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
2 -5 3.1250000001
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two complex roots, i.e. the roots are no longer real for (k < -3.125):
x = 1.25 + (0.0000070710681)i
x = 1.25 - (0.0000070710681)i
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Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
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Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
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Solve systems of linear equations up to 4-equations 4-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
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