Question

Refer to Problem 54. Show that $B^2-4AC$is invariant.

Short answer

So,$B^2-4AC$is invariant.

Step-by-step solution

Step 1. Given Information

We need to show that$B^2-4AC$is invariant.

Step 2. Explanation to invariant

From Q. 54

The equation is $x^2+4xy+4y^2+5\sqrt{5}y+5=0$

So, $B^2-4AC=4^2-4\times 1\times 5=-4$

The rotated equation is $0x{\text{'}}^2+5y{\text{'}}^2-2.3\sqrt{5}x\text{'}+4.4\sqrt{5}y\text{'}+5=0$

$B^2-4AC=0^2-4\times 1\times 1=-4$

Both results are equal.