Mathematics Skill Development - Module 2 Mathematics Skill Development - Module 2 Radical Expressions and Equations The following questions will evaluate the student's ability to manipulate radical expressions and solve radical equations 1. Express 2+V5 so that the denominator contains no radicals. 4-V5 Solution: The difference of squares formula is very useful to eliminate unwanted radicals. In this case: 2 + V5 2 + V5 4 + V5 4 - V5 4 - V5 4 + V5 = 8+615+5 16 - 5 = 13+61/5 . 11 2. Simplify 3/64a6. Solution: 3/64a6 = ((2a)6)1/3 = (((2a)2)3)1/3 = (2a)2 = 4a2 . q p xxyx. 3. Simplify Solution: r q 1/2) 1/2 ) 1/2 x = =(x.x3/4)1/2 = (x7/4) 1/2 = x78. 4. Simplify v48 + 175 - v/108. Solution: 1/48+ 175- v108 = V42.3+v52.3-162.3 = 4V3+5V3-6V3 = 31/3 = 33/2 1
Mathematics Skill Development - Module 2 5. Solve x + 4 = Vx + 10. Solution: We first square both sides of the given equation, we then compute x + 4 = Vx + 10 (x+4)2 = (Vx + 10)2 (x+4)(x+4) = x + 10 x2+8x+16=x+ 10 x2+7x+6=0 (x+6)(x+1) = 0. Thus, x = - 6 and x = - 1 are two possible solutions. When we square both sides of an equation, we must be careful to check that we have not introduced ny spurious solutions. In this case, when x = - 6, -6+4= - 2+2=v-6+10 and x = - 6 does not solve the original equation. However, when x = - 1, -1+4=3=v-1+10. Therefore x = - 1 is the only solution to the original equation. 6. Solve 2x - 1 - Vx - 4 = 2. Solution: Rearranging the equation and squaring both sides: 2x - 1 = 2 + Vx - 4 (2x - 1)2 = (2 + Vx - 4)2 2x-1=4+4vx-4+x-4 x - 1 = 4x - 4. Squaring both sides of the last equation gives (x-1)2 = (4Vx-4)2 22 - 2x +1 =16(x-4) x2-18x+65=0 (x - 13)(x- 5) = 0. Therefore x = 13 and x = 5 are possible solutions to this equation. As before, when we square both sides of an equation, it is necessary to check that we haven't introduced any spurious solutions. When x = 13, and when x = 5, 12 . 13 - 1 - v13 - 4 = 5 -3 =2 2
Mathematics Skill Development - Module 2 V2.5-1-v5-1=3-1=2 so that both x = 13 and x = 5 are solutions of our original equation. 7. Find two integers a and b such that a+ v2 1 b - V2 V2 (Hint: What happens when b = 2?). Solution: If b = 2, rationalizing the denominator gives a + V2 a+ v2 2+12 · b- V2 2- 12 2+ 12 = 2a+(2+a)v2+2 4-2 = 2(a+1)+(2+a)v/2 2 1 12 . =(a+1)+ (2+ a). If we choose a = - 1, the last expression is just 1. So if we choose b = 2, we must set a