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I look at math homework
(1) Prove: △=[-(k+2)] -4×2k squared.

=(k+2) -8k squared

= the square of k-4k+4

= the square of k-2

Square of (k-2) ≥0

No matter what real number k takes, the equation always has a real number root.

(2) A root of the equation = (k+2)+(k-2) = k.

The other root of the equation =2 (k+2)-(k-2)=2.

Because one side of the isosceles triangle ABC is A > 1, and a=2.

So the perimeter of triangle ABC

c=k+2+2

=k+4

This question is correct. Please refer to. You use symbols to represent some words because I can't. )