Question

prove that x^4+px^3+rx+s^2 is a perfect square, if ps=+-r and p^2+8s=0. hence solve x^4-4x^3+8x+4=0.​

Answer 1

Step-by-step explanation:

Let us suppose that x^4 +px^3+qx^2+rx+s=(x^2+ax+b)^2,since coefficient of x^4 is 1,coefficient of x^2 in (x^2+bx+c) is 1.

So, x^4 +px^3+qx^2+rx+s=x^4+2x^2(ax+b)+(ax+b)^2

=x^4+2ax^3+2bx^2+(a^2)x^2+2axb+b^2

Comparing LHS and RHS,

p=2a; q=2b+a^2; r=2ab; s=b^2

4q=8b+4a^2=8b+p^2,or,b=(4q-p^2)/8 and a=p/2

rs=2ab^3=p((4q-p^2)/8)^3