Find the constant k so that the quadratic equation 2x^2 + 5x - k = 0 has two real solutions
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Given the quadratic equation f(x) = 2x^2 + 5x -k
We need to find the value of k such that f(x) has two real solutions.
We will use delta to determine the values of k.
We know that the function has two real solution is delta > 0
delta = b^2 - 4ac
We know that: a = 2 b= 5 c = -k
==> b^2 - 4ac = (25 - 4*2*-k ) > 0
==> 25 + 8k > 0
==> 8k > -25
==> k > - 25/8
Then the values of k belongs to the interval ( -25/8 , inf)
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to find the value of k so that the quadratic equation 2x^2 + 5x - k = 0 has two real solutions.
Now the roots of a quadratic equation are given by [-b+ sqrt(b^2 - 4ac]/2a and [-b- sqrt(b^2 - 4ac]/2a.
For the roots to be real and different b^2 - 4ac > 0
=> 5^2 - 4*-k*2 > 0
=> 25 + 8k > 0
=> 8k > -25
=> 8k > -25/8
=> k > -25/8
Therefore k should be greater than -25/8.
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