# Find the constant k so that the quadratic equation 2x2 + 5x - k = 0 has two real solutions.

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

2x2 + 5x - k = 0

If the function has two real solutions , then delta must be a positive number greater than 0:

We know that:

delta = b^2 - 4*a*c > 0

Such that:

a = 2   b = 5     c = -k

Let us calculate:

delta = 5^2 - 4*2*-k  > 0

==> 25 + 8k > 0

Now subtract 25 from both sides:

==> 8k > -25

Now we will divide by 8:

==> k > -25/ 8

Then k values shoul bs greater that - 25/8 in order for the function to have two real roots.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To fond k such that 2x^2+5x-k = 0 has real solutions.

2x^2+5x -k = 0.

We divide both side of the equation by 2:

x^2 +(5/2)x -  k/2 = 0.

We add k/2 to both sides:

x^2-(5/2)x = k/2.

We add  (5/2)^2 to both sides so that left side is a perfect square .

x^2-(5/2)x +(5/2) = k/2+(5/2)^2.

(x-5/2)^2 = k/2.+(5/2)^2.

x-5/2 = +or- sqrt(k/2+25/4)

Now to have two  real distinct solutions, k/2 + 25/4 > 0.

Or k/2 > -25/4

Or k > -25/4 = -12.5.

Therefore  k > = -12.5 in order that there are two real solutions .

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

For the given equation to have 2 real solution, we'll put the discriminant delta to be positive or equal to zero.

delta > = 0

delta = b^2 - 4ac, where a,b,c are the coefficients of the quadratic.

Since it is not specified if the roots are different or they are equal, we'll consider both cases:

delta > 0

b^2 - 4ac > 0

We'll identify a,b,c:

2x^2 + 5x - k = 0

a = 2 , b = 5 , c = -k

delta = 25 + 8k

25 + 8k > 0

8k > -25

k > -25/8

So, when the values of k are in the interval (-25/8 ; +infinite), the quadratic has 2 different real roots.

For delta = 0 => x1 = x2

25 + 8k = 0

8k = -25

k = -25/8

For k = -25/8, the roots of the quadratic are equal and real.

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