Indices and surds questions and answers pdf printable form pdf template

[Pages:4]Indices and surds questions and answers pdf printable form pdf template

Indices and surds questions and answers pdf printable form pdf template

In addition and subtraction of surds we will learn how to find the sum or difference of two or more surds only when they are in the simplest form of like surds.For addition and subtraction of surds, we have to check the surds that if they are similar surds or dissimilar surds.Follow the following steps to find the addition and subtraction of two or more surds:Step I: Convert each surd in its simplest mixed form.Step II: Then find the sum or difference of rational co-efficient of like surds.Step III: Finally, to get the required sum or difference of like surds multiply the result obtained in step II by the surd-factor of like surds.Step IV: The sum or difference of unlike surds is expressed in a number of terms by connecting them with positive sign (+) or negative (-) sign. If the surds are similar, then we can sum or subtract rational coefficients to find out the result of addition or subtraction.\(a\sqrt[n]{x}\pm b\sqrt[n]{x} = (a\pm b)\sqrt[n]{x}\)The above equation shows the rule of addition and subtraction of surds where irrational factor is \(\sqrt[n]{x}\) and a, b are rational coefficients.Surds firstly need to be expressed in their simplest form or lowest order with minimum radicand, and then only we can find out which surds are similar. If the surds are similar, we can add or subtract them according to the rule mentioned above.For example we need to find the addition of \(\sqrt[2]{8}\), \(\sqrt[2] {18}\).Both surds are in same order. Now we need find express them in their simplest form.So \(\sqrt[2]{8}\) = \(\sqrt[2]{4\times 2}\) = \(\sqrt[2]{2^{2}\times 2}\) = \(2\sqrt[2]{2}\)And \(\sqrt[2]{18}\) = \(\sqrt[2]{9\times 2}\) = \(\sqrt[2]{3^{2}\times 2}\) = \(3\sqrt[2]{2}\).As both surds are similar, we can add their rational co-efficient and find the result. Now \(\sqrt[2]{8}\) + \(\sqrt[2]{18}\) = \(2\sqrt[2]{2}\) + \(3\sqrt[2]{2}\) = \(5\sqrt[2]{2}\).Similarly we will find out subtraction of \(\sqrt[2]{75}\), \(\sqrt[2]{48}\).\(\sqrt[2]{75}\)= \(\sqrt[2]{25\times 3}\)= \(\sqrt[2]{5^{2}\times 3}\)= \(5\sqrt[2]{3}\)\(\sqrt[2]{48}\) = \(\sqrt[2]{16\times 3}\) = \(\sqrt[2]{4^{2}\times 3}\)= \(4\sqrt[2]{3}\)So \(\sqrt[2]{75}\) - \(\sqrt[2]{48}\) = \(5\sqrt[2]{3}\) - \(4\sqrt[2]{3}\) = \(\sqrt[2]{3}\).But if we need to find out the addition or subtraction of \(3\sqrt[2]{2}\) and \(2\sqrt[2]{3}\), we can only write it as \(3\sqrt[2]{2}\) + \(2\sqrt[2]{3}\) or \(3\sqrt[2]{2}\) - \(2\sqrt[2]{3}\). As the surds are dissimilar, further addition and subtraction are not possible in surd forms. Examples of Addition and Subtraction of Surds:1. Find the sum of 12 and 27. Solution: Sum of 12 and 27 = 12 + 27 Step I: Express each surd in its simplest mixed form; = \(\sqrt{2\cdot 2\cdot 3}\) + \(\sqrt{3\cdot 3\cdot 3}\) = 23 + 33 Step II: Then find the sum of rational co-efficient of like surds. = 532. Simplify \(3\sqrt[2] {32}\) + \(6\sqrt[2]{45}\) - \(\sqrt[2]{162}\) - \(2\sqrt[2]{245}\).Solution:\(3\sqrt[2]{32}\) + \(6\sqrt[2]{45}\) - \(\sqrt[2]{162}\) - \(2\sqrt[2]{245}\)= \(3\sqrt[2]{16\times 2}\) + \(6\sqrt[2]{9\times 5}\) - \(\sqrt[2]{81\times 2}\) - \(2\sqrt[2]{49\times 5}\)= \(3\sqrt[2]{4^{2}\times 2}\) + \(6\sqrt[2]{3^{2}\times 5}\) - \(\sqrt[2]{9^{2}\times 2}\) - \ (2\sqrt[2]{7^{2}\times 5}\)= \(12\sqrt[2]{2}\) + \(18\sqrt[2]{5}\) - \(9\sqrt[2]{2}\) - \(14\sqrt[2]{5}\)= \(3\sqrt[2]{2}\) + \(4\sqrt[2]{5}\) 3. Subtract 245 from 420. Solution: Subtract 245 from 420 = 420 - 245 Now convert each surd in its simplest form = 4\(\sqrt{2\cdot 2\cdot 5}\) - 2\(\sqrt{3\cdot 3\cdot 5}\) = 85 - 65 Clearly, we see that 85 and 65 are like surds. Now find the difference of rational co-efficient of like surds = 25. 4. Simplify \(7\sqrt[3]{128}\) + \(5\sqrt[3]{375}\) - \(2\sqrt[3]{54}\) - \(2\sqrt[3]{1029}\).Solution:\(7\sqrt[3]{128}\) + \(5\sqrt[3]{375}\) - \(2\sqrt[3]{54}\) - \(2\sqrt[3]{1029}\)= \(7\sqrt[3]{64\times 2}\) + \(5\sqrt[3]{125\times 3}\) - \(\sqrt[3]{27\times 2}\) - \(2\sqrt[3]{343\times 3}\)= \(7\sqrt[3]{4^{3}\times 2}\) + \(5\sqrt[3]{5^{3}\times 3}\) - \(\sqrt[3]{3^{3}\times 2}\) - \(2\sqrt[3]{7^{3}\times 3}\)= \(28\sqrt[3]{2}\) + \(25\sqrt[3]{3}\) - \(3\sqrt[3]{2}\) - \(14\sqrt[3]{3}\)= \(25\sqrt[3]{2}\) + \(11\sqrt[3]{3}\). 5. Simplify: 58 - 2 + 550 - 2\(^{5/2}\) Solution: 58 - 2 + 550 - 2\(^{5/2}\) Now convert each surd in its simplest form = 5\(\sqrt{2\cdot 2\cdot 2}\) - 2 + 5\(\sqrt{2\cdot 5\cdot 5}\) - \(\sqrt{2^{5}}\) = 5\(\sqrt{2\cdot 2\cdot 2}\) - 2 + 5\(\sqrt{2\cdot 5\cdot 5}\) - \(\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2}\) = 102 - 2 + 252 - 42 Clearly, we see that 85 and 65 are like surds. Now find the sum and difference of rational co-efficient of like surds = 302 6. Simplify \(24\sqrt[3]{3}\) + \(5\sqrt[3]{24}\) - \(2\sqrt[2]{28}\) - \(4\sqrt[2]{63}\).Solution:\(24\sqrt[3]{3}\) + \(5\sqrt[3]{24}\) - \(2\sqrt[2]{28}\) - \(4\sqrt[2]{63}\)= \(24\sqrt[3]{3}\) + \(5\sqrt[3]{8\times 3}\) - \(2\sqrt[2]{4\times 7}\) - \(4\sqrt[2]{9\times 7}\)= \(24\sqrt[3]{3}\) + \(5\sqrt[3]{2^{3}\times 3}\) - \(2\sqrt[2] {2^{2}\times 7}\) - \(4\sqrt[2]{3^{2}\times 7}\)= \(24\sqrt[3]{3}\) + \(10\sqrt[3]{3}\) - \(4\sqrt[2]{7}\) - \(12\sqrt[2]{7}\)= \(34\sqrt[3]{3}\) - \(16\sqrt[2]{7}\). 7. Simplify: 25 - 54 + 316 - 625 Solution: 25 - 54 + 316 - 625 Now convert each surd in its simplest form = 25 - \(\sqrt[3]{2\cdot 3\cdot 3\cdot 3}\) + 3\(\sqrt[3]{2\cdot 2\cdot 2\cdot 2}\) - \(\sqrt[3]{5\cdot 5\cdot 5\cdot 5}\) = 25 - 32 + 62 - 55 = (62 - 32) + (25 - 55), [Combining the like surds] Now find the difference of rational co-efficient of like surds = 32 - 35 8. Simplify \(5\sqrt[2]{7}\) + \(3\sqrt[2]{20}\) - \(2\sqrt[2]{80}\) - \(3\sqrt[2]{84}\).Solution:\(5\sqrt[2]{7}\) + \(3\sqrt[2]{20}\) - \(2\sqrt[2]{80}\) - \ (3\sqrt[2]{84}\)= \(5\sqrt[2]{7}\) + \(3\sqrt[2]{4\times 5}\) - \(2\sqrt[2]{16\times 5}\) - \(3\sqrt[2]{16\times 6}\)= \(5\sqrt[2]{7}\) + \(3\sqrt[2]{2^{2}\times 5}\) - \(2\sqrt[2]{4^{2}\times 2}\) - \(3\sqrt[2]{4^{2}\times 6}\)= \(5\sqrt[2]{7}\) + \(6\sqrt[2]{5}\) - \(8\sqrt[2]{5}\) - \(12\sqrt[2]{6}\)= \(5\sqrt[2]{7}\) - \(2\sqrt[2]{5}\) - \(12\sqrt[2] {6}\).Note: x + y \(\sqrt{x + y}\) and x - y \(\sqrt{x - y}\) Surds 11 and 12 Grade Math From Addition and Subtraction of Surds to HOME PAGE Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. Share this page: What's this? GCSE 6 7AQAEdexcelOCRWJECHigherAQA 2022Edexcel 2022OCR 2022WJEC 2022 Level 6-7 GCSE The fractional indices laws apply when the power is a fraction. \textcolor{red}{a}^{\large{\frac{\textcolor{blue}{b}}{\textcolor{limegreen}{c}}}} = \sqrt[\textcolor{limegreen}{c}]{\textcolor{red}{a}^\textcolor{blue}{b}} This is commonly use to show square and cube roots. \textcolor{red}{x}^{\large{\frac{\textcolor{limegreen}{1}}{\textcolor{blue}{2}}}}= \sqrt[\textcolor{blue}{2}]{\textcolor{red}{x}^\textcolor{limegreen}{1}} =\sqrt[\textcolor{blue}{2}]{\textcolor{red}{x}} \textcolor{red}{x}^{\large{\frac{\textcolor{limegreen}{1}}{\textcolor{blue}{3}}}}= \sqrt[\textcolor{blue} {3}]{\textcolor{red}{x}^\textcolor{limegreen}{1}} =\sqrt[\textcolor{blue}{3}]{\textcolor{red}{x}} Note: it doesn't matter which order you carry out the square root and multiplication operations. In other words, the rule can also be written as \textcolor{red}{a}^{\large{\frac{\textcolor{blue}{b}}{\textcolor{limegreen}{c}}}} = (\sqrt[\textcolor{limegreen}{c}]{\textcolor{red}{a}})^\textcolor{blue}{b} You should try to carry out the operations in the order that makes the calculation as simple as possible. Level 6-7 GCSE You may also be asked to simplify expressions where the numerator is not \bf{1}. \textcolor{red}{64}^{\large{\frac{\textcolor{limegreen}{2}} {\textcolor{blue}{3}}}}= \sqrt[\textcolor{blue}{3}]{\textcolor{red}{64}^\textcolor{limegreen}{2}} \sqrt[\textcolor{blue}{3}]{\textcolor{red}{64}} = \textcolor{red}{4} \textcolor{red}{4}^\textcolor{limegreen}{2} = \textcolor{red}{16} Level 6-7 GCSE Negative powers flip the fraction and put 1 over the number In general, the result of a negative power is "\bf{1} over that number to the positive power", i.e. \textcolor{red}{a}^{-\textcolor{limegreen}{b}} = \dfrac{1}{\textcolor{red}{a}^\textcolor{limegreen}{b}} for any value of a or b. When the power is \textcolor{blue}{-1}, this takes the form, \textcolor{red}{a}^{\textcolor{blue}{-1}}=\dfrac{1}{\textcolor{red}{a}} or \textcolor{red}{10}^{\textcolor{blue}{-1}} = \dfrac{1}{\textcolor{red}{10}} When the number is a fraction, the negative power flips the fraction. \bigg(\dfrac{\textcolor{blue}{a}}{\textcolor{limegreen}{b}}\bigg)^{-\textcolor{red}{x}} = \bigg(\dfrac{\textcolor{limegreen}{b}}{\textcolor{blue}{a}}\bigg)^\textcolor{red}{x} Level 6-7 GCSE Level 6-7 GCSE Simplify the following, 4^{-3}. [2 marks] We now know that 4^{-3} is equal to \dfrac{1}{4^3}. We also know that 4^3=4\times 4\times 4=16\times 4=64. So, we get that 4^{-3}=\frac{1}{64}. Level 6-7 GCSE Simplify the following, 9^{\frac{3}{2}}. [2 marks] So, we know that 9^{\frac{3}{2}} is equal to \sqrt[2]{9^3} or (\sqrt[2]{9})^3. So, to work out (\sqrt[2]{9})^3, we first have to square root 9, which is easy enough ? the square root of 9 is 3. So, (\sqrt[2]{9})^3 becomes 3^3, which is 3^3=3\times 3\times 3 = 27 Level 6-7 GCSE Write 2^{15}\times 8^{-4} as a power of 2, and hence evaluate the expression. (Non calculator) [3 marks] The first part of the expression is a power of 2, whilst the second part is a power of 8. we know that 8 = 2^3 This means we can rewrite the following, 8^{-4}=\left(2^3\right)^{-4} Next, using Rule 3, we can simplify, \left(2^3\right)^{-4}=2^{3\times(-4)}=2^{-12} So the whole expression can be written as 2^{15}\times2^{-12}, Finally using Rule 1 we simplify the expression further. 2^{15}\times2^{-12}=2^{15+(-12)}=2^3 Thus, we have written the expression as a power of 2. Evaluating this final answer gives 2^3 = 8 Level 6-7 GCSE Example Questions So, we can't use any laws straight away since the terms don't have the same base. However, if we recognise that 9=3^2, then we can write the first term as \left(3^2\right)^5 Using the power law, we get \left(3^2\right)^5=3^{2\times5}=3^{10} Therefore, the whole expression becomes 3^{10}\times3^{-5} Applying the multiplication law, this simplifies to 3^{10+(-5)}=3^5 Thus, we have written the expression as a power of 3. Firstly, as 3^2=9, the inverse operation gives, \sqrt{9}=3 So, that leaves 6^{-2}, this becomes the following fraction, 6^{-2}=\dfrac{1}{6^2} We know that 6^2=6\times 6=36, so 6^{-2}=\dfrac{1}{36} Multiplying our two answers together, we get \sqrt{9}\times 6^{-2}=3\times\dfrac{1}{36}=\dfrac{3}{36}=\dfrac{1}{12} This expression can be rewritten as, \sqrt4 \times (\sqrt4)^3 Given we know that \sqrt4=2 , this becomes, 2\times2^3 Hence, 2\times2^3=2\times8=16 Notice that in this example we chose to perform the \sqrt{4} operation before cubing the answer. We could alternatively write the expression as \sqrt{4^3}, but in this case the first option is easier. As it is a negative power we can rewrite this as, 8^{-\frac{5}{3}}=\frac{1} {8^{\frac{5}{3}}} Now, we can work out the denominator, which we will write as, 8^{\frac{5}{3}}=\sqrt[3]{8^5}=(\sqrt[3]{8})^5 We know that \sqrt[3]{8}=2. So this simplifies to, (\sqrt[3]{8})^5=2^5 Counting up in powers of 2: 4, 8, 16, 32 ? we see that 32 is the 5th power of 2, so \sqrt[3]{8}^5=32 Therefore, the answer is, 8^{-\frac{5} {3}}=\dfrac{1}{32} Related Topics Worksheet and Example Questions Drill Questions You May Also Like... Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC. ?8.99 View Product The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. 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