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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 + 5)(x – Factor the difference of two squares, x2 – 25. %3D x + 5 = or Set each factor equal to 0. X- X = -5 Solve each linear equation. %3D Check each solution by substituting it into the original equation. Check -5: Check 5: x² – 25 = 0 x2 – 25 = 0 )2 - 25 ?= 0 52 – 25 ?= 0 - 25 ?= 0 25 – 25 ?= 0 = 0 True 0 = 0 True The solutions of x2 – 25 are and 5 and the solution set is {-5, 5}. Solve: x2 - 169 = 0 X = (smaller value) X = (larger value)

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 + 5)(x – Factor the difference of two squares, x2 – 25. %3D x + 5 = or Set each factor equal to 0. X- X = -5 Solve each linear equation. %3D Check each solution by substituting it into the original equation. Check -5: Check 5: x² – 25 = 0 x2 – 25 = 0 )2 - 25 ?= 0 52 – 25 ?= 0 - 25 ?= 0 25 – 25 ?= 0 = 0 True 0 = 0 True The solutions of x2 – 25 are and 5 and the solution set is {-5, 5}. Solve: x2 - 169 = 0 X = (smaller value) X = (larger value)

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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I need help with all of them plz. 

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 + 5)(x – Factor the difference of two squares, x² – 25. %D x + 5 or x - Set each factor equal to 0. %3D X = -5 Solve each linear equation. = Check each solution by substituting it into the original equation. Check -5: Check 5: x2 - 25 = 0 x2 – 25 = 0 )2 – 25 ?= 0 52 - 25 ?= 0 25 ?= 0 25 – 25 ?= 0 - = 0 True 0 = 0 True The solutions of x² – 25 are and 5 and the solution set is {-5, 5}. Solve: x2 169 = 0 X = (smaller value) X = (larger value)
Transcribed Image Text: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 + 5)(x – Factor the difference of two squares, x² – 25. %D x + 5 or x - Set each factor equal to 0. %3D X = -5 Solve each linear equation. = Check each solution by substituting it into the original equation. Check -5: Check 5: x2 - 25 = 0 x2 – 25 = 0 )2 – 25 ?= 0 52 - 25 ?= 0 25 ?= 0 25 – 25 ?= 0 - = 0 True 0 = 0 True The solutions of x² – 25 are and 5 and the solution set is {-5, 5}. Solve: x2 169 = 0 X = (smaller value) X = (larger value)
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