2.2 Circles 1 of 2 Decide what numbers must be added to both sides of the equation to make the left side the sum of a perfect square trinomial in \( x \) and a perfect square trinomial in \( y \). \( x^{2}+24 x+y^{2}-20 y=-4 \) The number \( \square \) should be added to both sides of the equation to make the left side the sum of a perfect square trinomial in \( x \). (Simplify your answer.)

To make the left side of the equation a sum of perfect square trinomials in \( x \) and \( y \), we need to complete the square for both \( x \) and \( y \). For \( x \), we have \( x^2 + 24x \). To complete the square, we need to add the square of half the coefficient of \( x \), which is \( (24/2)^2 = 12^2 = 144 \). For \( y \), we have \( y^2 - 20y \). To complete the square, we need to add the square of half the coefficient of \( y \), which is \( (-20/2)^2 = (-10)^2 = 100 \). Now, let's add these numbers to both sides of the original equation: \( x^2 + 24x + 144 + y^2 - 20y + 100 = -4 + 144 + 100 \) This simplifies to: \( (x^2 + 24x + 144) + (y^2 - 20y + 100) = 140 \) The left side of the equation is now a sum of perfect square trinomials: \( (x + 12)^2 + (y - 10)^2 = 140 \) So, the number that should be added to both sides of the equation to make the left side the sum of a perfect square trinomial in \( x \) is \( 144 \).