Solve for all values of x by factoring. x^(2)-23=2 Answer: x=

Add 23: Add 23 to both sides of the equation to isolate the x^2 term. We have the equation x^2 - 23 = 2. To solve for x, we first want to isolate the x^2 term on one side of the equation. We do this by adding 23 to both sides of the equation. x^2 - 23 + 23 = 2 + 23 x^2 = 25Take square root: Take the square root of both sides of the equation to solve for x. Now that we have x^2 = 25, we can take the square root of both sides to solve for x. Remember that taking the square root of a number gives us two solutions: one positive and one negative. √(x^2) = ±√(25) x = ±5Check solutions: Check the solutions in the original equation. We have found that x = ±5. We should substitute these values back into the original equation to ensure they are correct. For x = 5: (5)^2 - 23 = 25 - 23 = 2, which is true. For x = -5: (-5)^2 - 23 = 25 - 23 = 2, which is also true. Both solutions satisfy the original equation.

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Solve for all values of
x by factoring.

x^(2)-23=2
Answer:
x=

Solve for all values of $x$ by factoring. $\newline$ $x^{2}-23=2$ $\newline$ Answer: $x=$

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Math Problems

Algebra 1

Factor quadratics with leading coefficient 1

Full solution

Q. Solve for all values of

x

by factoring.

\newline

x^{2}-23=2

\newline

Answer:

x=

Add $23$ : Add $23$ to both sides of the equation to isolate the $x^2$ term. $\newline$ We have the equation $x^2 - 23 = 2$ . To solve for $x$ , we first want to isolate the $x^2$ term on one side of the equation. We do this by adding $23$ to both sides of the equation. $\newline$ $x^2 - 23 + 23 = 2 + 23$ $\newline$ $x^2 = 25$
Take square root: Take the square root of both sides of the equation to solve for $x$ . $\newline$ Now that we have $x^2 = 25$ , we can take the square root of both sides to solve for $x$ . Remember that taking the square root of a number gives us two solutions: one positive and one negative. $\newline$ $\sqrt{x^2} = \pm\sqrt{25}$ $\newline$ $x = \pm5$
Check solutions: Check the solutions in the original equation. $\newline$ We have found that $x = \pm 5$ . We should substitute these values back into the original equation to ensure they are correct. $\newline$ For $x = 5$ : $\newline$ $(5)^2 - 23 = 25 - 23 = 2$ , which is true. $\newline$ For $x = -5$ : $\newline$ $(-5)^2 - 23 = 25 - 23 = 2$ , which is also true. $\newline$ Both solutions satisfy the original equation.