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

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

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

Answer

Explanation:

Step1: Complete the square

For the equation (x^{2}+10x = 24), the coefficient of (x) is (10). Half of it is (\frac{10}{2}=5), and then square it ((\frac{10}{2})^{2}=25). Add (25) to both sides of the equation: (x^{2}+10x + 25=24 + 25) Using the formula ((a + b)^{2}=a^{2}+2ab + b^{2}) (here (a=x), (b = 5)), the left - hand side becomes ((x + 5)^{2}). So, ((x + 5)^{2}=49)

Step2: Solve for (x)

Take the square root of both sides: (x+5=\pm\sqrt{49}=\pm7) When (x + 5=7), then (x=7 - 5=2) When (x + 5=-7), then (x=-7 - 5=-12)

Answer:

({-12,2})