微差事

4 个 =2,3^{4}=1=3$）个^{44} = = "4 , 2^3= 8\)。 所以 \text{a}_{n+1} = \frac{a_n}{2} + 1\)， The given quadratic equation is: $$ (x+1)(x-5) + (x-3)(x+2) = 0 $$ First, let's expand both products in the equation. 1. Expand $(x+1)(x-5)$: $$ (x+1)(x-5) = x^2 - 5x + x - 5 = x^ to find the roots of the equation $(x+1)(x-5) + (x-3)(z+2) = 0$. It seems like there was a mistake in your question as it got cut off at the end. However, I'll assume you want to solve the equation: $$ (x+1)(x-5) + (x-3)(x+2) = 0 $$ ### Step 1: Expand both products **First product:** $$ (x+1)(x-5) = x^2 - 5x + x - 5 = x^2 - 4x - 5 $$ **Second product:** $$ (x-3)(x+2) = x^2 + 2x - 3x - 6 = x^2 - x - 6 $$ ### Step 2: Add the two expanded expressions Now, add the two results together: $$ (x^2 - 4x - 5) + (x^2 - x - 6) $$ Combine like terms: $$ x^2 + x^2 - 4x - x - 5 - 6 = 2x^2 - 5x - 11 $$ So the equation becomes: $$ 2x^2 - 5x - 11 = 0 $$ ### Step 3: Solve the quadratic equation We now solve the quadratic equation: $$ 2x^2 - 5x - 11 = 0 $$ Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, $ a = 2 $, $ b = -5 $, and $ c = -11 $. Plug these into the formula: $$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-11)}}{2(2)} $$ Simplify inside the square root: $$ x = \frac{5 \pm \sqrt{25 + 88}}{4} = \frac{5 \pm \sqrt{113}}{4} $$ Since $\sqrt{113}$ is irrational, the roots are: $$ x = \frac{5 \pm \sqrt{113}}{4} $$ ### Final Answer: The solutions to the equation are: $$ x = \frac{5 \pm \sqrt{113}}{4} $$