How do we find out the discriminant for this quadratic equation?(a+c)ax2+2bx+c=2(ac−b2)(x2+1)If I bring the RHS to LHS and then expand , it will take a lot of time for me to first expand and then find the discriminant for this equation whence is the first way to find the discriminant. Is there a quicker way with which I can find its discriminant ? Maybe just by looking at the equation , not even have to expand.

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ncruuk7ikt · Accepted Answer

You can tell at a glance that the RHS has no linear term in x. You can also tell at a glance that the quadratic coefficient on the RHS is&amp;nbsp;2ac−2b2. The quadratic coefficient on the LHS is&amp;nbsp;a2+ca, and if you subtract the LHS from the RHS, the resulting quadratic coefficient is&amp;nbsp;a2+2b2−ac. The linear coefficient is just what it is on the LHS, since the RHS has no linear term, so that is 2b. The constant term is also told at a glance; it is c on the LHS and&amp;nbsp;2ac−2b2.The discriminant is&amp;nbsp;y2−4xz, where x,y,z are the quadratic, linear and constant coefficients respectively.x=a2+2b2−ac,y=2b,z=c+2b2−2ac. Slightly messily, we compute the discriminant as:4b2−4(a2+2b2−ac)(c+2b2−2ac)=4b2−4(a2c+2a2b2−2a3c+2b2c+4b4−2acb2−ac2−2acb2+2a2c2)&amp;nbsp;=4b2−4(4b4+2a2b2+2a2c2+a2c+2b2c−2a3c−4acb2−ac2)&amp;nbsp;=4b2+16acb2+4ac2+8a3c−16b4−8a2b2−8a2c2−4a2c−8b2c&amp;nbsp;=4b2(1+4ac−4b2−2a2−2c)+4a(c2+2a2c−2ac2−ac)

How do we find out the discriminant for

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