As per the given question,
x3−25x2−4x+100=0,x^3-25x^2-4x+100=0,x3−25x2−4x+100=0,
Now,
x3−4x−25x2+100=0,x^3-4x-25x^2+100=0,x3−4x−25x2+100=0,
x(x2−4)−25(x2−4)=0,x(x^2-4)-25(x^2-4)=0,x(x2−4)−25(x2−4)=0,
(x2−4)(x−25)=0,(x^2-4)(x-25)=0,(x2−4)(x−25)=0,
so,
x2=4x^2=4x2=4 or x=25x=25x=25,
x=±2x=\pm2x=±2 or x=25x=25x=25.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Cauchy-Schwarz inequality does not directly produce a solution of the problem. It just shows that x^2-4=k(x-25), where k is a constant, x^2-kx+25k-4=0. The discriminant is D=k^2-4(25k-4) and it should be a square of a rational number.
Your answer is not showing Cauchy schwarz equality
Comments
Cauchy-Schwarz inequality does not directly produce a solution of the problem. It just shows that x^2-4=k(x-25), where k is a constant, x^2-kx+25k-4=0. The discriminant is D=k^2-4(25k-4) and it should be a square of a rational number.
Your answer is not showing Cauchy schwarz equality