solve for x in the equation $x^{2}-10x + 25 = 35$.\n$x = 5pm2sqrt{5}$\n$x = 5pmsqrt{35}$\n$x =…

solve for x in the equation $x^{2}-10x + 25 = 35$.\n$x = 5pm2sqrt{5}$\n$x = 5pmsqrt{35}$\n$x = 10pm2sqrt{5}$\n$x = 10pmsqrt{35}$

solve for x in the equation $x^{2}-10x + 25 = 35$.\n$x = 5pm2sqrt{5}$\n$x = 5pmsqrt{35}$\n$x = 10pm2sqrt{5}$\n$x = 10pmsqrt{35}$

Answer

Explanation:

Step1: Recognize perfect - square trinomial

The left - hand side of the equation $x^{2}-10x + 25$ is a perfect - square trinomial. It can be factored as $(x - 5)^{2}$. So the equation becomes $(x - 5)^{2}=35$.

Step2: Take square root of both sides

Taking the square root of both sides, we get $x−5=\pm\sqrt{35}$.

Step3: Solve for x

Add 5 to both sides of the equation: $x = 5\pm\sqrt{35}$.

Answer:

B. $x = 5\pm\sqrt{35}$