solve $x^{2}-8x = 3$ by completing the square. which is the solution set of the equation?\n{4 - \\sqrt{19}…

solve $x^{2}-8x = 3$ by completing the square. which is the solution set of the equation?\n{4 - \\sqrt{19}, 4 + \\sqrt{19}}\n{4 - \\sqrt{11}, 4 + \\sqrt{11}}\n{4 - \\sqrt{8}, 4 + \\sqrt{8}}\n{4 - \\sqrt{3}, 4 + \\sqrt{3}}

solve $x^{2}-8x = 3$ by completing the square. which is the solution set of the equation?\n{4 - \\sqrt{19}, 4 + \\sqrt{19}}\n{4 - \\sqrt{11}, 4 + \\sqrt{11}}\n{4 - \\sqrt{8}, 4 + \\sqrt{8}}\n{4 - \\sqrt{3}, 4 + \\sqrt{3}}

Answer

Explanation:

Step1: Rewrite the equation

Given $x^{2}-8x = 3$. The coefficient of $x$ is $- 8$. Half of it is $\frac{-8}{2}=-4$, and its square is $(-4)^{2}=16$. Add 16 to both sides of the equation: $x^{2}-8x + 16=3 + 16$.

Step2: Factor the left - hand side

The left - hand side is a perfect square trinomial. Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = x$ and $b = 4$, we have $(x - 4)^{2}=19$.

Step3: Solve for x

Take the square root of both sides: $x-4=\pm\sqrt{19}$. Then $x = 4\pm\sqrt{19}$.

Answer:

{4 - $\sqrt{19}$, 4 + $\sqrt{19}$}