×(×+5)=2
(s+6)²=15
(×+2)²+(×–3)²=9
(2r+3)²+(r+4)²=10

Answer:Let's solve each of these equations step-by-step:1. **Equation 1: x(x + 5) = 2** Expand and simplify:\[ x^2 + 5x = 2\]\[ x^2 + 5x - 2 = 0\] This is a quadratic equation. Use the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a = 1, b = 5, and c = -2:\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\]\[ x = \frac{-5 \pm \sqrt{25 + 8}}{2}\]\[ x = \frac{-5 \pm \sqrt{33}}{2}\] So, the solutions are:\[ x = \frac{-5 + \sqrt{33}}{2} \quad \text{and} \quad x = \frac{-5 - \sqrt{33}}{2}\]2. **Equation 2: (s + 6)^2 = 15** Take the square root of both sides:\[ s + 6 = \pm \sqrt{15}\] Solve for s:\[ s = -6 \pm \sqrt{15}\]3. **Equation 3: (x + 2)^2 + (x - 3)^2 = 9** Expand and simplify:\[ (x + 2)^2 = x^2 + 4x + 4\]\[ (x - 3)^2 = x^2 - 6x + 9\]\[ x^2 + 4x + 4 + x^2 - 6x + 9 = 9\]\[ 2x^2 - 2x + 13 = 9\]\[ 2x^2 - 2x + 4 = 0\]\[ x^2 - x + 2 = 0\] This quadratic equation has no real roots because the discriminant (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 is negative.4. **Equation 4: (2r + 3)^2 + (r + 4)^2 = 10** Expand and simplify:\[ (2r + 3)^2 = 4r^2 + 12r + 9\]\[ (r + 4)^2 = r^2 + 8r + 16\]\[ 4r^2 + 12r + 9 + r^2 + 8r + 16 = 10\]\[ 5r^2 + 20r + 25 = 10\]\[ 5r^2 + 20r + 15 = 0\]\[ r^2 + 4r + 3 = 0\] Factorize the quadratic:\[ (r + 1)(r + 3) = 0\] So, the solutions are:\[ r = -1 \quad \text{and} \quad r = -3\]In summary:1. x = \frac{-5 \pm \sqrt{33}}{2}2. s = -6 \pm \sqrt{15}3. No real solutions4. r = -1 and r = -3

Answered by esteabapo | 2024-09-09