for what value of k are the roots of the quadratic equation (k+4)x*x+(k+1)x+1=0?
To find the value of k for which the given quadratic equation has roots, we can use the quadratic formula. The quadratic formula states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the quadratic equation is (k+4)x^2 + (k+1)x + 1 = 0.
By comparing this with the general quadratic equation form, we can see that a = k+4, b = k+1, and c = 1.
Substituting these values into the quadratic formula, we get:
x = (-(k+1) ± √((k+1)^2 - 4(k+4)(1))) / (2(k+4))
Now, for the equation to have roots, the discriminant (b^2 - 4ac) must be greater than or equal to zero. So, we have:
(k+1)^2 - 4(k+4)(1) ≥ 0
Expanding and simplifying this inequality:
k^2 + 2k + 1 - 4k - 16 ≥ 0
k^2 - 2k - 15 ≥ 0
Now, we solve this quadratic inequality:
Factorizing the left side:
(k - 5)(k + 3) ≥ 0
The roots of this inequality are k = -3 and k = 5, which divide the number line into three intervals: (-∞, -3), (-3, 5), and (5, ∞).
To determine for which interval the inequality is true, we can choose test points from each interval:
For k < -3, let's choose k = -4:
(-4 - 5)(-4 + 3) ≥ 0
(-9)(-1) ≥ 0
9 ≥ 0 (true)
For -3 < k < 5, let's choose k = 0:
(0 - 5)(0 + 3) ≥ 0
(-5)(3) ≥ 0
-15 ≥ 0 (false)
For k > 5, let's choose k = 6:
(6 - 5)(6 + 3) ≥ 0
(1)(9) ≥ 0
9 ≥ 0 (true)
From this, we see that the inequality is true for k < -3 and k > 5.
Therefore, the value of k for which the roots of the quadratic equation (k+4)x^2 + (k+1)x + 1 = 0 exist is k < -3 or k > 5.