The equation (k+3)x^2 + 6x + k =5 has two distinct real solutions for x. Prove that k^2-2k-24<0
We argue using the discriminant of a quadratic polynomial. For a quadratic ax^2 + bx + c=0, the discriminant D is D = b^2 - 4ac. When this quadratic has two distinct real roots, we have that D > 0. So we proceed by putting "0" on one side of the equation and plugging in the values from our expression. Here in our case, a = (k+3), b= 6, c=(k-5). (Notice c is not equal to k, since we need to have 0 on one side of the equation and so subtract 5 from both sides first).62-4(k+3)(k-5)>0Squaring 6 and multiplying out the brackets we get:36-4(k2+3k-5k-15)>0Collecting terms we get:36-4(k2-2k-15)>0Now we can divide everything by 4 to get:9-(k2-2k-15)>0Putting everything over to the other side we get:k2-2k-24>0And we are done.
