University:
Jackson College
Course:
MAT-010 | Basic Mathematics
Academic year:

2024
Views:

385

Pages:

4
Author:

Bennett Wheeler

Question 1 Express each of the following in simplest radical form. All variables represent positive real numbers. 5 12 𝑥 4 12x 4 5 Answer 1 We have to express in simplest radical form: 5 12 𝑥 4 12x 4 5 Using the property of square roots: 5 12 𝑥 4 = 5 12 𝑥 4 12x 4 5 = 12x 4 5 Simplify the denominator: Rewrite 12 𝑥 4 12x 4 as 4 ⋅ 3 ⋅ 𝑥 4 4⋅3⋅x 4 : = 5 2 𝑥 2 3 = 2x 2 3 5 So, the simplest form of the expression is: 5 2 𝑥 2 3 2x 2 3 5 Question 2 Rationalize the denominator and simplify. All variables represent positive real numbers. 3 𝑦 3 2 𝑥 − 3 𝑦 3 2 x −3 3 y 3 3 y Answer 1 To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator: 3 𝑦 3 2 𝑥 − 3 𝑦 3 ⋅ 2 𝑥 + 3 𝑦 3 2 𝑥 + 3 𝑦 3 2 x −3 3 y 3 3 y ⋅ 2 x +3 3 y 2 x +3 3 y Applying the difference of squares in the denominator: = 3 𝑦 3 ( 2 𝑥 + 3 𝑦 3 ) ( 2 𝑥 ) 2 − ( 3 𝑦 3 ) 2 = 6 𝑥 𝑦 3 + 9 𝑦 3 2 4 𝑥 − 9 𝑦 2 / 3 = (2 x ) 2 −(3 3 y ) 2 3 3 y (2 x +3 3 y ) = 4x−9y 2/3 6 x 3 y +9 3 y 2 The simplified form is: 6 𝑥 𝑦 3 + 9 𝑦 3 2 4 𝑥 − 9 𝑦 2 / 3 4x−9y 2/3 6 x 3 y +9 3 y 2 Question 3 Determine the value of the variable for which the expression is defined as a real number. ( 1 𝑥 2 − 2 𝑥 − 63 ) 1 3 ( x 2 −2x−63 1 ) 3 1 Answer 1 Since we have a cube root, the expression is defined for values of 𝑥 x such that 𝑥 2 − 2 𝑥 − 63 ≠ 0 x 2 −2x−63  =0. Factor the quadratic: 𝑥 2 − 2 𝑥 − 63 = ( 𝑥 − 9 ) ( 𝑥 + 7 ) = 0 x 2 −2x−63=(x−9)(x+7)=0 Solve for 𝑥 x: 𝑥 = 9 or 𝑥 = − 7 x=9orx=−7 Thus, the expression is defined for all 𝑥 x except 𝑥 ≠ 9 x  =9 and 𝑥 ≠ − 7 x  =−7. Question 4 Add and subtract radical expressions: 12 + 3 12 + 3 Answer 1 Simplify 12 12 : 12 = 4 ⋅ 3 = 2 3 12 = 4⋅3 =2 3 Add 12 12 and 3 3 : 2 3 + 3 = 3 3 2 3 + 3 =3 3 So, the answer is: 3 3 3 3 Question 5 Solve the radical equation and check all proposed solutions. 𝑥 − 3 𝑥 − 2 = 4 x− 3x−2 =4 Answer 1 Rearrange terms: 𝑥 − 4 = 3 𝑥 − 2 x−4= 3x−2 Square both sides: ( 𝑥 − 4 ) 2 = 3 𝑥 − 2 (x−4) 2 =3x−2 Expand and simplify: 𝑥 2 − 8 𝑥 + 16 = 3 𝑥 − 2 x 2 −8x+16=3x−2 𝑥 2 − 11 𝑥 + 18 = 0 x 2 −11x+18=0 Factor the quadratic: ( 𝑥 − 9 ) ( 𝑥 − 2 ) = 0 (x−9)(x−2)=0 So, 𝑥 = 9 x=9 or 𝑥 = 2 x=2. Check each solution: For 𝑥 = 9 x=9: 9 − 3 ⋅ 9 − 2 = 9 − 27 − 2 = 9 − 25 = 9 − 5 = 4 9− 3⋅9−2 =9− 27−2 =9− 25 =9−5=4 𝑥 = 9 x=9 is a solution. For 𝑥 = 2 x=2: 2 − 3 ⋅ 2 − 2 = 2 − 6 − 2 = 2 − 4 = 2 − 2 = 0 2− 3⋅2−2 =2− 6−2 =2− 4 =2−2=0 𝑥 = 2 x=2 does not satisfy the original equation. Therefore, the only solution is: 𝑥 = 9 x=9