show that if k is any real number, the quadratic equation k^2+(3k+2)x+(2k+3)=0 always has real roots

Question

this is a thinking and inquiry question, I have no clue how to tackle it. I assum you use a discriminant, but I need help, Thanks

Answer 1

The quadratic variable is k, rather than x. The equation can be written k^2+3kx+2k+(2x+3)=0 or k^2+k(3x+2)+(2x+3)=0. Using the quadratic formula we can see that the discriminant is (3x+2)^2-4(2x+3)=9x^2+12x+4-8x-12=9x^2+4x-8, another quadratic. Because the constant term is negative, the discriminant of this quadratic is 16+4*9*8, which is positive, and positive numbers have rational roots. Therefore the quadratic in k will also have real roots, even if they are irrational.

If k is a real number and the variable or unknown is x, then we can write x=-(k^2+2k+3)/(3k+2) and this is a simple linear equation in x, and its value is real because k is real. However, x would not be defined when the denominator is 0, when 3k+2=0 or k=-2/3.