Complete the table. Use long pad as your answer sheet. No. Given Quadratic Equation 1. $x(x+5) = 2$ 2. $(s+6)^2 = 15$ 3. $(t-2)^2 + (t-3)^2 = 9$ 4. $(2r + 3)^2 + (r + 4)^2 = 10$ 5. $(m-4)^2+(m-7)^2=15$ a b c $X_1$ $X_2$ $X_1 + X_2$ $X_1 \times X_2$
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Then we square 12, which equals 144. So, X1 = 144. For Equation 2: (-2)^2 = 4 We square -2, which equals 4. So, X2 = 4. For Equation 3: (-3)^2 = 9 We square -3, which equals 9. So, X3 = 9. For Equation 4: (2r + 3)^2 + (4)^2 = 10 We need to solve for r in this Show more…
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