'Solve: x2 _ 25 = 0 We will factor the binomial on the left side of the equation and use the zero-factor property. To use the zero-factor property, we need one side of the equation to be factored completely and the other side to be 0. We factor the difference of two squares on the left side of the equation and proceed as follows x2 _ 25 This is the equation to solve: (x + s)(x - O Factor the difference of two squares, 25 x + 5 = X-L Set each factor equal to 0. -5 Solve each linear equation. Check each solution by substituting it into the original equation_ Check -5: Check 5: X2 _ 25 = x2 _ 25 )2 25 ?= 0 52 _ 25 ? = 0 25 ? = 0 25 25 ? = 0 = 0 True = 0 True The solutions of x2 25 are and 5 and the solution set is {~5, 5} Solve: x2 169 = 0 (smaller value) (larger value)'
Added by Nathaniel R.
Step 1
We can factor this as a difference of squares: $(x + 5)(x - 5) = 0$. Now, we can set each factor equal to 0: $x + 5 = 0$ and $x - 5 = 0$. Solving these equations, we get $x = -5$ and $x = 5$. So, the solution set is $\boxed{\{-5, 5\}}$. Show more…
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