Question

For what value of k are there two distinct real solutions to the original quadratic equation (k+1)x²+4kx+2=0.

Answer 1

Answer:

k ∈ (-∞,$-\frac{1}{2}$)∪(1,∞)

Step-by-step explanation:

For quadratic equations $ax^2+bx+c=0,a\neq 0$ you can find the solutions with the Bhaskara's Formula:

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

A quadratic equation usually has two solutions.

If you only want real solutions the condition is that the discriminant ($\Delta$) has to be greater than zero, this means:

$\Delta=b^2-4ac>0$

Then we have the expression:

$(k+1)x^2+4kx+2=0$

$a=(k+1)\\b=4k\\c=2$

Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:

$b^2-4ac>0\\(4k)^2-4.(k+1).2>0\\16k^2-8(k+1)>0\\16k^2-8k-8>0$

We got another quadratic function.

$16k^2-8k-8>0$ we can simplify the expression dividing both sides in 8.

$16k^2-8k-8>0\\\\\frac{16k^2}{8} -\frac{8k}{8} -\frac{8}{8} >\frac{0}{8}\\\\2k^2-k-1>0$

We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.

$2k^2-k-1>0\\a=2\\b=-1\\c=-1$

$k_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1+\sqrt{9} }{4}=\frac{1+3}{4} =1 \\and\\k_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$

Then,

$k>1 \\and\\k$

The answer is:

For all the real values of k who belongs to the interval:

(-∞,$-\frac{1}{2}$)∪(1,∞)

there are two distinct real solutions to the original quadratic equation $(k+1)x^2+4kx+2=0$