# The roots of the equation 5x^2 - kx + 1 = 0 are real and distinct. Calculate all posible values of k.

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### 3 Answers

5x^2 - kx +1 =0

if the roots are real, then te dicriminant should be positive.

==> delta = b^2 - 4ac > 0

==> k^2 - 4*5*1 > 0

==> k^2 - 20 > 0

==> (k-sqrt20)(k+sqrt2) > 0

==> (k - 2sqrt5) (k+sqrt5)> 0

==> k > 2sqrt5 and k > -2sqrt5

==> K = (2sqrt5, inf)

OR:

==> k < 2sqrt5 and K < -2sqrt5

==> k = (-inf, - 2sqrt5)

==> K belongs to (-inf, -2sqrt5) U (2sqrt5, inf)

5x^2-kx+1 = 0. To find the condition for real and distinct roots.

Solution

LHS

5x^2 -kx+1 = 5(x^2-kx/5)+1

= 5(x^2- kx/5 + (k/10)^2) - 5(k/10)^2 +1

= 5(x - k/10)^2 - (5k^2-100)/10^2.

So the given euation could be written as:

5(x-k/10)^2 = (5k^2-100)/10^2.

If the roots are distinct , then right side should be real and positive as left side is a perfect square.

So 5k^2-100 > 0.

5k^2 > 100

k^2 = 20

k > sqrt 20 = 2sqrt5. Or

k < -s2sqrt5.

The roots of the equation 5x^2 - kx + 1 = 0 are real and distinct, when the discriminant of the equation, delta, will be positive.

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

Let's identify a,b,c:

a = 5

b = -k

c = 1

delta = (k)^2 - 4*1*5

delta > 0 => k^2 - 20 > 0

k^2 - 20 = 0

We'll add 20 both sides:

k^2 =20

k1 = +sqrt 20

k2 = -sqrt 20

k1 = +2sqrt5

k2 = -2sqrt5

Delta is positive and the equation has 2 distinct real roots when k is in the intervals:

**(-inf., -2sqrt5) U (2sqrt5,+inf.)**