solve $x^{2}+10x = 24$ by completing the square. which is the solution set of the equation?\n$\\{-5…

solve $x^{2}+10x = 24$ by completing the square. which is the solution set of the equation?\n$\\{-5 - \\sqrt{34},-5 + \\sqrt{34}\\}$\n$\\{-5 - \\sqrt{29},-5 + \\sqrt{29}\\}$\n$\\{-12,2\\}$\n$\\{-2,12\\}$

solve $x^{2}+10x = 24$ by completing the square. which is the solution set of the equation?\n$\\{-5 - \\sqrt{34},-5 + \\sqrt{34}\\}$\n$\\{-5 - \\sqrt{29},-5 + \\sqrt{29}\\}$\n$\\{-12,2\\}$\n$\\{-2,12\\}$

Answer

Explanation:

Step1: Add square of half - coefficient of x.

The coefficient of $x$ is $10$. Half of it is $5$ and its square is $25$. Add $25$ to both sides of the equation $x^{2}+10x = 24$. So we get $x^{2}+10x + 25=24 + 25$.

Step2: Rewrite left - hand side as a perfect square.

The left - hand side $x^{2}+10x + 25=(x + 5)^{2}$, and the right - hand side is $49$. So, $(x + 5)^{2}=49$.

Step3: Take square root of both sides.

$x+5=\pm\sqrt{49}=\pm7$.

Step4: Solve for x.

When $x + 5=7$, $x=7 - 5=2$. When $x + 5=-7$, $x=-7 - 5=-12$.

Answer:

C. ${-12,2}$