Prove that the roots of the equation kx^2+(2k+4)x+8=0 are real for all values of k.
For a quadratic of the form ax^2 + bx + c = 0 to have real roots,
the discriminant b^2 - 4ac ≥ 0
(2k+4)^2 - 4k(8) ≥ 0
4k^2 + 16k + 16 - 32k ≥ 0
k^2 - 4k + 4 ≥ 0
(k-2)^2 ≥ 0
(k - 2) ≥ 0
k-2 ≥ 0 OR -k+2 ≥ 0
k ≥ 2 OR -k ≥ -2 ----> k ≤ 2
Well, k≥ 2 OR k≤2 would refer to all values of k
Recall the discriminant, b^2-4ac
If it is positive, there are two real roots.
So, is (2k+4)^2 - 4*k*8 always positive?
as (k-2) is squared its always positive...so the discriminant is always greater than zero.???
Why did the mathematician go broke? He started squaring the circle, and all his money just went round and round! But hey, I'd be happy to help you out!
To prove that the roots of the equation are real for all values of k, we can use the discriminant (b² - 4ac) of the quadratic formula. In this case, the equation is kx² + (2k+4)x + 8 = 0.
Let's calculate the discriminant: (2k+4)² - 4(k)(8). Simplify that and you'll get: 4k² + 16k + 16 - 32k. Combining like terms, we have: 4k² - 16k + 16.
Now let's simplify further: 4(k² - 4k + 4). Notice that inside the parentheses, we have a perfect square trinomial: (k-2)². So the discriminant can be written as: 4(k-2)².
Since 4 is always positive and (k-2)² is always non-negative (since it's being squared), the discriminant is always greater than or equal to zero. And since the discriminant being greater than or equal to zero is a condition for having real roots, we've proven that the equation has real roots for all values of k. Ta-da!
To prove that the roots of the equation \(kx^2 + (2k+4)x + 8 = 0\) are real for all values of \(k\), we need to demonstrate that the discriminant of the quadratic equation is non-negative.
The discriminant (\(D\)) of a quadratic equation \(ax^2 + bx + c = 0\) can be calculated using the formula: \(D = b^2 - 4ac\). If the discriminant is positive (\(D>0\)), the equation has two distinct real roots. If the discriminant is zero (\(D=0\)), the equation has one real root (a repeated root). And if the discriminant is negative (\(D<0\)), the equation has two complex conjugate roots.
In this case, the equation is \(kx^2 + (2k+4)x + 8 = 0\). Comparing this with the standard form \(ax^2 + bx + c = 0\), we have \(a = k\), \(b = 2k+4\), and \(c = 8\).
Now, let's calculate the discriminant (\(D\)) for this equation:
\[D = (2k+4)^2 - 4(k)(8)\]
\[D = 4k^2 + 16k + 16 - 32k\]
\[D = 4k^2 - 16k + 16\]
To prove that the roots are real for all values of \(k\), we need to show that \(D \geq 0\) for all \(k\). In other words, we need to prove that \(4k^2 - 16k + 16 \geq 0\) for all \(k\).
To solve this inequality, we can attempt to find the values of \(k\) that make the inequality true. One approach is to factorize the expression if possible. For this particular inequality, however, it is easier to solve it by using the quadratic formula.
The quadratic formula states that if a quadratic equation \(ax^2 + bx + c = 0\) has real roots, they can be calculated using the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our case, the quadratic equation is \(4k^2 - 16k + 16 = 0\). By comparing it with the standard form \(ax^2 + bx + c = 0\), we can see that \(a = 4\), \(b = -16\), and \(c = 16\).
Now let's use the quadratic formula to find the roots:
\[k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(4)(16)}}{2(4)}\]
\[k = \frac{16 \pm \sqrt{256 - 256}}{8}\]
\[k = \frac{16 \pm \sqrt{0}}{8}\]
\[k = \frac{16 \pm 0}{8}\]
\[k = \frac{16}{8}\]
\[k = 2\]
From this, we can observe that the quadratic equation has a single real root (a repeated root) when \(k = 2\). Since there is only one value of \(k\) that makes the discriminant zero, we can conclude that for all other values of \(k\) (except \(k = 2\)), the discriminant is always positive (\(D>0\)). Therefore, the roots of the equation \(kx^2 + (2k+4)x + 8 = 0\) are real for all values of \(k\) except \(k = 2\).