Chapter 1 LP 1-1 (a) A va of the real-worls variables. p. 5 LP 1-2 (a) to fin operations in the ddition or subtri P 1-3 (a) 2[(3n P 1-4 20.5 ; 6 1-5 (a) They $r$ wer: $\frac{1}{5}$ is the 1 procal is alwa $s$ are rearrang he terms are $65(7+2) ; 4$ heck (1) d $.4+1.6+i$ ution p. 16 , tion p. 16, 551 p. 3. . 42-43) Let's Practice 9-7. Solve each equation using the square root property. a. [tex]x^2=4[/tex] b. [tex](x+1)^2=7[/tex] c. [tex](4 x+3)^2=25[/tex] d. [tex]6 x^2=54[/tex] e. [tex]x^2=\frac{1}{4}[/tex] f. [tex](3 h-2)^2=4[/tex] Solve each equation by factoring. g. [tex]x^2-4 x+3=0[/tex] h. [tex]2 k^2-40=-11 k[/tex] i. [tex]16 y^3=25 y[/tex] j. [tex]x^2-4 x=5[/tex] k. [tex]\frac{9}{25} m^2=121[/tex] l. [tex]-7-18 x+9 x^2=0[/tex]

Use the square root property to solve equations of the form x 2 = a , resulting in x = ± a ​ .
Factor quadratic equations of the form a x 2 + b x + c = 0 to find solutions.
For x 2 = 4 , the solutions are x = ± 2 .
For x 2 − 4 x + 3 = 0 , the solutions are x = 1 and x = 3 .
The solutions to the equations are: a) x = ± 2 , b) x = − 1 ± 7 ​ , c) x = − 2 , 2 1 ​ , d) x = ± 3 , e) x = ± 2 1 ​ , f) h = 0 , 3 4 ​ , g) x = 1 , 3 , h) k = − 8 , 2 5 ​ , i) y = 0 , ± 4 5 ​ , j) x = − 1 , 5 , k) m = ± 3 55 ​ , l) x = − 3 1 ​ , 3 7 ​ .

Explanation

Introduction We are given a set of equations to solve, some using the square root property and others by factoring. Let's tackle them one by one!

Solving a a. x 2 = 4 To solve this using the square root property, we take the square root of both sides: x = ± 4 ​ x = ± 2 So, x = 2 or x = − 2 .

Solving b b. ( x + 1 ) 2 = 7 Taking the square root of both sides: x + 1 = ± 7 ​ x = − 1 ± 7 ​ So, x = − 1 + 7 ​ or x = − 1 − 7 ​ .

Solving c c. ( 4 x + 3 ) 2 = 25 Taking the square root of both sides: 4 x + 3 = ± 25 ​ 4 x + 3 = ± 5 We have two cases: Case 1: 4 x + 3 = 5 4 x = 2 x = 4 2 ​ = 2 1 ​ Case 2: 4 x + 3 = − 5 4 x = − 8 x = − 2 So, x = 2 1 ​ or x = − 2 .

Solving d d. 6 x 2 = 54 Divide both sides by 6: x 2 = 6 54 ​ = 9 Taking the square root of both sides: x = ± 9 ​ x = ± 3 So, x = 3 or x = − 3 .

Solving e e. x 2 = 4 1 ​ Taking the square root of both sides: x = ± 4 1 ​ ​ x = ± 2 1 ​ So, x = 2 1 ​ or x = − 2 1 ​ .

Solving f f. ( 3 h − 2 ) 2 = 4 Taking the square root of both sides: 3 h − 2 = ± 4 ​ 3 h − 2 = ± 2 We have two cases: Case 1: 3 h − 2 = 2 3 h = 4 h = 3 4 ​ Case 2: 3 h − 2 = − 2 3 h = 0 h = 0 So, h = 3 4 ​ or h = 0 .

Solving g g. x 2 − 4 x + 3 = 0 We need to factor the quadratic. We are looking for two numbers that multiply to 3 and add to -4. These numbers are -1 and -3. ( x − 1 ) ( x − 3 ) = 0 So, x − 1 = 0 or x − 3 = 0 x = 1 or x = 3 .

Solving h h. 2 k 2 − 40 = − 11 k Rearrange the equation: 2 k 2 + 11 k − 40 = 0 We need to factor this quadratic. We are looking for two numbers that multiply to 2 ∗ ( − 40 ) = − 80 and add to 11. These numbers are 16 and -5. 2 k 2 + 16 k − 5 k − 40 = 0 2 k ( k + 8 ) − 5 ( k + 8 ) = 0 ( 2 k − 5 ) ( k + 8 ) = 0 So, 2 k − 5 = 0 or k + 8 = 0 2 k = 5 or k = − 8 k = 2 5 ​ or k = − 8 .

Solving i i. 16 y 3 = 25 y Rearrange the equation: 16 y 3 − 25 y = 0 Factor out y: y ( 16 y 2 − 25 ) = 0 y ( 4 y − 5 ) ( 4 y + 5 ) = 0 So, y = 0 or 4 y − 5 = 0 or 4 y + 5 = 0 y = 0 or 4 y = 5 or 4 y = − 5 y = 0 or y = 4 5 ​ or y = − 4 5 ​ .

Solving j j. x 2 − 4 x = 5 Rearrange the equation: x 2 − 4 x − 5 = 0 We need to factor this quadratic. We are looking for two numbers that multiply to -5 and add to -4. These numbers are -5 and 1. ( x − 5 ) ( x + 1 ) = 0 So, x − 5 = 0 or x + 1 = 0 x = 5 or x = − 1 .

Solving k k. 25 9 ​ m 2 = 121 25 9 ​ m 2 − 121 = 0 ( 5 3 ​ m ) 2 − 1 1 2 = 0 ( 5 3 ​ m − 11 ) ( 5 3 ​ m + 11 ) = 0 So, 5 3 ​ m − 11 = 0 or 5 3 ​ m + 11 = 0 5 3 ​ m = 11 or 5 3 ​ m = − 11 m = 3 5 ​ ∗ 11 or m = 3 5 ​ ∗ − 11 m = 3 55 ​ or m = − 3 55 ​ .

Solving l l. − 7 − 18 x + 9 x 2 = 0 Rearrange the equation: 9 x 2 − 18 x − 7 = 0 We need to factor this quadratic. We are looking for two numbers that multiply to 9 ∗ ( − 7 ) = − 63 and add to -18. These numbers are -21 and 3. 9 x 2 − 21 x + 3 x − 7 = 0 3 x ( 3 x − 7 ) + 1 ( 3 x − 7 ) = 0 ( 3 x + 1 ) ( 3 x − 7 ) = 0 So, 3 x + 1 = 0 or 3 x − 7 = 0 3 x = − 1 or 3 x = 7 x = − 3 1 ​ or x = 3 7 ​ .

Conclusion We have solved all the given equations using the square root property and factoring.


Examples
Understanding how to solve equations using the square root property and factoring is essential in many real-world applications. For instance, when designing structures, engineers use quadratic equations to model the behavior of materials under stress. Factoring these equations helps determine critical points where the structure might be at risk of failure. Similarly, in physics, these techniques are used to analyze projectile motion, calculating trajectories and impact points. Mastering these algebraic methods provides a foundation for solving complex problems in various scientific and engineering fields. For example, if you are designing a bridge, you might use these skills to calculate the tension and compression forces acting on different parts of the structure. Let's say the force can be modeled by x 2 − 4 x + 3 = 0 , where x represents the force. By solving this equation, you can find the forces at which the structure is stable.

Answered by GinnyAnswer | 2025-07-03

We solved the equations using the square root property and factoring techniques. Each method provided a clear pathway to the solutions for the given problems. This approach not only helps in mathematics but is also crucial in various applications and fields.
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Answered by Anonymous | 2025-07-04