I understand that these equations are equivalent but I do not understand how. Please show me how you manipulate the original equation to get the other equations. Help Thanks.

Original Equation:$\displaystyle x^4-13x^2+36=0$

$\displaystyle (2x^2)-2(2x^2)13+144=0$

$\displaystyle (2x^2)^2-2(2x^2)13+169=25$

Work it out, for example the first one:
$\displaystyle (2x)^2-2(2x^2)13+144=0$
$\displaystyle 4x^2-4x^2\cdot 13+144=0$
Dividing the ride and left side by 4 gives:
$\displaystyle x^2-13x^2+36=0$
which is the given equation.

For the other, put 25 from the right side to the left side and so: 169-25=144, which is equivalent to the other two equations.

Hello, theloser!

I understand that these equations are equivalent, but I do not understand how.
Please show me how you manipulate the original equation to get the other equation.

Original Equation: $\displaystyle x^4-13x^2+36\:=\:0$

$\displaystyle (2x^2)^2-2(2x^2)13+169=25$

We are given: . . . .$\displaystyle x^4 - 12x^2 + 36 \:=\:0$

Multiply by 4: . . .$\displaystyle 4x^4 - 52x^2 + 144 \:=\:0$

. . . . .$\displaystyle (2x^2)(2x^2) - 2\!\cdot\!2\!\cdot\!13\!\cdot\!x^2 + 144 \:=\:0$

. . . . . . . $\displaystyle (2x^2)^2 - 2(2x^2)(13) + 144 \:=\:0$

Add 25: .$\displaystyle (2x^2)^2 - 2(2x^2)(13) + 169 \:=\:25$

If they are trying to solve for x,
. . they're making a mess of it!