Questions and Answers #6 Algebra Foundations

Questions and Answers Sheet 6 Algebra Foundations Question #1 To determine: 6 11 increased by a number Answer: The given pharase in algebraic notation can be written as 6 increased by a number 11 6 = increassed by 𝑥 11 6 = plus 𝑥 11 6 = +𝑥 11 6 Thus, the algebraic expression for the given phrase is +𝑥 11 Question #2 Convert the given phrase into an algebraic expression 1 The quotient of the a number and − 5 Answer: To write an algebraic expression for a given phrase, first assume the number is 'x' 1 So is: x is subtracted from − 5 This is where the keywords are subtracted, which tells us that this is a subtraction operation. We now search for words from within a phrase to determine which ones are subtracted. 1 In this expression, we see that the terms x are subtracted from − 5 So, this phrase in algebraic notation can be written as 1 \text{The quotient of the a number and − 5 1 = The quotient of x and − 5 1 = 𝑥 divided by − 5 1 = 𝑥 ÷ (− ) 5 1 Thus, the algebraic expression for the given phrase is 𝑥 ÷ (− ) 5 Question #3 To determine: To translate given into an algebraic expression using x to represent the number 2 The given pharase is −5 decreased by a number 7 Answer: Let us suppose the number be x 2 2 The phrase −5 decreased by a number means decrease the number x from −5 7 The following algebraic expression will be formed2 −5 − 𝑥 7 7 =− 37 7 −𝑥 2 The algebraic expression is −5 − 𝑥 7 Question #4 To translate the given phrase into algebraic expression 2 of a number 3 Answer: In order to write the algebraic expression for the given phrase, first let's assume a number is 'x' 2 We have of a number, and here th ekey word is 'of', which tells us the operation is multiplication. 3 So the given in algebraic notation can be written as 2 of a number 3 2 = of𝑥 3 2 ×𝑥 3 2 Thus, the algebraic expression for phrase '' of a number'' is: 3 2 3 ×𝑥 Question #5 Translate the given phrase into algebraic expression 1 2 Answer: Thus, we can rephrase the given phrase as: The sum of x and 1 2 Here the keyword is sum, which tells us the operation is addition Now we look for the words and in the phrase to identify the number to be added to expression we can see that the numbers that the numbers that is being added to 1 2 1 2 . In the given is x So the given phrase in algebraic notation can be written as 1 1 = Sum of x and 2 2 1 = x plus 2 1 = x+ 2 Sum of a number and Thus, the algebraic expression for the given phrase is 𝑥 + 1 2 Question #6 Translate the given phrase into algebraic expression 2 − 5 Answer: In order to write the algebraic expression for the given phrase, first let's assume that a number is'x' 2 Thus, we can rephrase the given phrase as: x incresed by − 5 Here the keyword is increased by, which tells us the operation is addition 2 Now we look for the words by in the phrase to identify the number to be added to − . In the given expression we can see that the numbers that is being added to − 2 5 5 is x So the given phrase in algebraic notation can be written as 2 2 = x incresed by − 5 5 2 = x plus − 5 2 = x + (− ) 5 A number increase by − 2 Thus, the algebraic expression for the given phrase is 𝑥 + (− ) 5 Question #7 Translate the given phrase into algebraic expression. 7 20 Answer: The difference of x and 7 20 . Here the keyword is difference, which tells us the operation is subtraction. Now we look for the words and in the phrase to identify the number to be subtracted. In the given 7 expression we can see that the numbers that is being subtracted from x is 20 So, 7 7 = Difference of x and 20 20 7 = x minus 20 7 = x− 20 The difference of a number and The algebraic expression for the given phrase is 𝑥 − 7 20 Question #8 Find the zeros (if any) of the function 𝑦 = 𝑥 4 + 3𝑥 2 + 2 algebraically. Answer: Step 1 Given equation is 𝑦 = 𝑥 4 + 3𝑥 2 + 2. Step 2 In order to find the zeroes of given equation putting y = 0 , then the solution of the equation will give the zeroes. Step 3 The given equation can be solved algebraically by firstly making factors of the expression and then equating each factor to zero as shown below: 𝑥 4 + 3𝑥 2 + 2 = 0 𝑥 4 + 𝑥 2 + 2𝑥 2 + 2 = 0 𝑥 2 (𝑥 2 + 1) + 2(𝑥 2 + 1) = 0 (𝑥 2 + 1)(𝑥 2 + 2) = 0 Step 4 Equating each factor to zero, it gives (𝑥 2 + 1)(𝑥 2 + 2) = 0 𝑥2 + 1 = 0 𝑥 = ±𝑖 and 𝑥 2 + 2 = 0 𝑥 = ±𝑖√2 Step 5 Thus, the given function has four imaginary zeroes only which are: 𝑥 = 𝑖, −𝑖, 𝑖√2, −𝑖√2 Question #9 Solve the equation algebraically. Verify your results using a graphing utility. 1 2 2 3 𝑥 4 + = Answer: Step 1 Given the equation 1 2 2 3 𝑥 4 + = , We need to solve the equation graphically. Step 2 1 2 3 2 𝑥 4 Let us write the given equation + = 1 subtract on both sides 2 1 1 2 3 1 − + = − 2 2 𝑥 4 2 2 3 1 = − 𝑥 4 2 multiply 4 on both sides 2 3 1 𝑥 4 2 4( ) = 4( − ) 8 𝑥 8 𝑥 = 3−2 =1 x=8 Question #10 27 = 35𝑥 ⋅ 9𝑥 2 Answer: Step 1 Consider the given equation 2 27 = 35𝑥 ⋅ 9𝑥 Step 2 Simplify the equation further as, 2 27 = 35𝑥 ⋅ (32 )𝑥 33 = 35𝑥 ⋅ 32𝑥 2 33 = 35𝑥+2𝑥 2 Step 3 Equate the exponents. 3 = 5𝑥 + 2𝑥 2 2𝑥 2 + 5𝑥 − 3 = 0 𝑥= −5±√52 −4(2)(−3) 1 4 𝑥 = , −3 2 Step 4 1 Thus, the value of x is 𝑥 = , −3 2 Question #11 Let a be a complex number that is algebraic over Q. Show that √𝑎 is algebraic over Q. Answer: Step 1 Consider the polynomial: ℎ(𝑥) = 𝑥 2 − 𝑎 Clearly, ℎ(√𝑎) = 0 Hence, we have following: [𝑄(√𝑎): 𝑄(𝑎)] ≤ 𝑑𝑒𝑔(ℎ) Let f(x) be the minimal polynomial in Q[x] Step 2 Now by tower rule, we have: [𝑄(√𝑎): 𝑄] = [𝑄(√𝑎): 𝑄(𝑎)][𝑄(𝑎): 𝑄] ≤ 𝑑𝑒𝑔 ℎ(𝑥). 𝑑𝑒𝑔 𝑓(𝑥) Hence, [𝑄(√𝑎): 𝑄] is a finite extension and so algebraic. Step 3 Answer: [𝑄(√𝑎): 𝑄] is algebraic. Question #12 Solve the following inequality algebraically. x +14x +40 <0 Answer: Step 1 Here the given inequality is 𝑥 2 + 14𝑥 + 40 ≤ 0 (1) Step 2 Let us factorize ′′𝑥 2 + 14𝑥 + 40′′ 𝑥 2 + 14𝑥 + 40 ⇒ 𝑥 2 + 10𝑥 + 4𝑥 + 40 ⇒ 𝑥(𝑥 + 10) + 4(𝑥 + 10) ⇒ (𝑥 + 10)(𝑥 + 4)(2) From (1) and (2) we get (𝑥 + 10)(𝑥 + 4) ≤ 0(3) From (3) we get, either 𝑥 + 10 ≤ 0 and 𝑥 + 4 ≥ 0 ⇒ 𝑥 ≤ −10 and 𝑥 ≥ −4 , which is impossible. Or, from (3) we get 𝑥 + 10 ≥ 0 and 𝑥 + 4 ≤ 0 ⇒ 𝑥 ≥ −10 and x ≤ −4 ⇒ 𝑥 ∈ [−10, −4] So, the required answer will be 𝑥 ∈ [−10, −4]. Question #13 An exponential function of the form 𝑦 = 𝑎(𝑏) 𝑥 passes through the points (2,1176) and (5, 50421). Determine the values of a and b algebraically. Answer: Graph An exponential function of the form 𝑦 = 𝑎(𝑏) 𝑥 passes through the points (2,1176) and (5, 50421). Determine the values of a and b algebraically. Solution 𝑦 = 𝑎𝑏 𝑥 passes through (2,1176) and (5,50421) 1176 = 𝑎𝑏 2 (1) 50421 = 𝑎𝑏 5 (2) Now from (1) we get, 1176 𝑎= 2 𝑏 (2) we get, 1176 50421 = 2 𝑏 5 𝑏 50421 3 =𝑏 1176 343 = 𝑏3 8 7 2 =𝑏 (1) we get, 49 1176 = 𝑎. 4 𝑎 = 96 7 Therefore the value of a=96 and 𝑏 = . 2 Question #14 Unit Name: Polynomial and Rational Functions Solve the following inequality algebraically. Explain your process. 8𝑥 − 3 ≤ 2𝑥 + 1 ≤ 17𝑥 − 8 Answer: Split the inequality into two: 8𝑥 − 3 ≤ 2𝑥 + 1 and 2𝑥 + 1 ≤ 17𝑥 − 8 Isolate x from each inequality using inverse operations. 8𝑥 − 3 ≤ 2𝑥 + 1 8𝑥 − 2𝑥 ≤ 3 + 1 6𝑥 ≤ 4 2 𝑥≤ 3 AND 2𝑥 + 1 ≤ 17𝑥 − 8 2𝑥 − 17𝑥 ≤ − 8 − 1 −15𝑥 + 15𝑥 ≤ −9 + 15𝑥 0 ≤ −9 + 15𝑥 3 ≤𝑥 5 Therefore, the solution is: 3 5 ≤𝑥≤ 2 3 Question #15 Solve the following system of linear equation algebraically. 2x+3y=2 x-4y=12 Answer: Step 1 the given system of the equation as 2x+3y=2 (1) x-4y=12 (2) Thus the equation (2) can be written as x=12+4y (3) Step 2 The value of equation (3) put in equation (1) as 2(12+4y)+3y=2 24+8y+3y=2 11y=2-24 11y=-22 22 𝑦=− 11 y=-2 Step 3 Now this value of y put in equation (3) as x=12+4y x=12+4.(-2) x=12-8 x=4 Hence the solution of the system is (x, y)=(4, -2) Question #16 Solve the following equation algebraically. Explain your process. 0 = 18𝑥 4 + 87𝑥 3 + 3𝑥 2 − 108𝑥 Answer: Step 1 Given equation is: 18𝑥 4 + 87𝑥 3 + 3𝑥 2 − 108𝑥 = 0 Taking 3x as common term we get, 3𝑥(6𝑥 3 + 29𝑥 2 + 𝑥 − 36) = 0 So either, 3x = 0 x=0 Or 6𝑥 3 + 29𝑥 2 + 𝑥 − 36 = 0 Step 2 Splitting up the second and third terms we get, 6𝑥 3 − 6𝑥 2 + 35𝑥 2 − 35𝑥 + 36𝑥 − 36 = 0 6𝑥 2 (𝑥 − 1) + 35𝑥(𝑥 − 1) + 36(𝑥 − 1) = 0 Taking (x - 1) as common term we get, (𝑥 − 1)(6𝑥 2 + 35𝑥 + 36) = 0 Using middle term factorization we get, (𝑥 − 1)(6𝑥 2 + 8𝑥 + 27𝑥 + 36) = 0 (𝑥 − 1)[2𝑥(3𝑥 + 4) + 9(3𝑥 + 4)] = 0 (𝑥 − 1)(3𝑥 + 4)(2𝑥 + 9) = 0 (𝑥 − 1) = 0; (3𝑥 + 4) = 0; (2𝑥 + 9) = 0 4 9 𝑥 = 1; 𝑥 = − ; 𝑥 = − 3 2 Hence the roots of the given equation is: 4 9 𝑥 = 0; 𝑥 = 1; 𝑥 = − ; 𝑥 = − . 3 2 Question #17 Try to solve for the indicated variable A=P+Prt Solve for P Answer: Step 1 Here we solve the equation A=P+Prt, solve for P Step 2 Here A=P+Prt ⇒ 𝑃 + 𝑃𝑟𝑡 = 𝐴 ⇒ 𝑃(1 + 𝑝𝑡) = 𝐴 ⇒𝑃= 𝐴 1+𝑟𝑡 Hence 𝑃 = 𝐴 1+𝑟𝑡 Question #18 Solve the equation algebraically. Verify your results using a graphing utility. 5(𝑥 − 2) − 2(3𝑥 + 1) = 4(1 − 2𝑥) + 𝑥 Answer: Given: 5(𝑥 − 2) − 2(3𝑥 + 1) = 4(1 − 2𝑥) + 𝑥 Calculation: 5(𝑥 − 2) − 2(3𝑥 + 1) = 4(1 − 2𝑥) + 𝑥 5𝑥 − 10 − 6𝑥 − 2 = 4 − 8𝑥 + 𝑥 -x-12=-7x+4 -x+7x=4+12 6x=16 16 𝑥= 𝑥= 6 8 3 Question #19 Find the zeros of the function algebraically. 𝑓(𝑥) = 2𝑥 2 − 3𝑥 − 54 Answer: Step 1 Given function is 𝑓(𝑥) = 2𝑥 2 − 3𝑥 − 54 To find the zeroes of the function. Solution: Finding zeroes of the given function. 2𝑥 2 − 3𝑥 − 54 = 0 2𝑥 2 − 12𝑥 + 9𝑥 − 54 = 0 2𝑥(𝑥 − 6) + 9(𝑥 − 6) = 0 (𝑥 − 6)(2𝑥 + 9) = 0 (x-6)=0 or (2x+9)=0 9 x=6 or 𝑥 = − 2 Step 2 9 Hence, zeroes of the given function are 𝑥 = {− , 6} 2 Question #20 Solve 𝑥 6 − 7𝑥 3 − 8 = 0 algebraically. Find real solutions only, if they exist. Answer: Put 𝑢 = 𝑥 3 then rewrite the equation 𝑢2 − 7𝑢 − 8 = 0 𝑢2 − 8𝑢 + 𝑢 − 8 = 0 𝑢(𝑢 − 8) + 1(𝑢 − 8) = 0 (𝑢 + 1)(𝑢 − 8) = 0 𝑢 + 1 = 0, 𝑢 − 8 = 0 u=-1 or 8u=8 Put 𝑢 = 𝑥 3 then solve x 𝑢 = −1 ⇒ 𝑥 3 = −1 ⇒ (𝑥 + 1)(𝑥 2 − 𝑥 + 1) = 0 ⇒ 𝑥 + 1 = 0 or 𝑥 2 − 𝑥 + 1 = 0 x=-1 or 𝑥 = +1±√1−4 2 1 ± √−3 𝑥= 2 1±√3𝑖 𝑥= 2 [𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, 𝑥 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 ] and u=8 𝑥 3 = 8 ⇒ 𝑥 3 = (2)3 ⇒ 𝑥 2 − (2)3 = 0 ⇒ (𝑥 − 2)(𝑥 2 + 2𝑥 + 4) = 0 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) ⇒ 𝑥 − 2 = 0 or 𝑥 2 + 2𝑥 + 2 = 0 x=2 or 𝑥 = −2±√4−16 2 −2 ± √−12 𝑥= 2 −2±√12𝑖 𝑥= 2 𝑥= −2±2√3𝑖 2 𝑥 = −1 ± √3𝑖 Hence real roots are -1 and 2.