MATH SOLVE

4 months ago

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

Accepted Solution

A:

Answer:k ∈ (-∞,[tex]-\frac{1}{2}[/tex])∪(1,∞)Step-by-step explanation:For quadratic equations [tex]ax^2+bx+c=0,a\neq 0[/tex] you can find the solutions with the Bhaskara's Formula:[tex]x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]A quadratic equation usually has two solutions.If you only want real solutions the condition is that the discriminant ([tex]\Delta[/tex]) has to be greater than zero, this means:[tex]\Delta=b^2-4ac>0[/tex]Then we have the expression:[tex](k+1)x^2+4kx+2=0[/tex][tex]a=(k+1)\\b=4k\\c=2\\[/tex]Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:[tex]b^2-4ac>0\\(4k)^2-4.(k+1).2>0\\16k^2-8(k+1)>0\\16k^2-8k-8>0[/tex]We got another quadratic function.[tex]16k^2-8k-8>0[/tex] we can simplify the expression dividing both sides in 8.[tex]16k^2-8k-8>0\\\\\frac{16k^2}{8} -\frac{8k}{8} -\frac{8}{8} >\frac{0}{8}\\\\2k^2-k-1>0[/tex]We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.[tex]2k^2-k-1>0\\a=2\\b=-1\\c=-1[/tex][tex]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}[/tex]Then,[tex]k>1 \\and\\k<-\frac{1}{2}[/tex]The answer is:For all the real values of k who belongs to the interval:(-∞,[tex]-\frac{1}{2}[/tex])∪(1,∞)there are two distinct real solutions to the original quadratic equation [tex](k+1)x^2+4kx+2=0[/tex]