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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/igcse/maths/edexcel/grade-7-7-1-surds/flashcards","label":"Surds","subjectCode":"igcse_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"b4fb5f75-1e89-4dc1-b1e3-70bdf6d29481","title":"Surds Flashcards","topicId":"93bf8b6f-a42a-4df9-a766-7beba6eb1d7f","topicSlug":"grade-7-7-1-surds","cards":[{"id":"2abcc871-0049-4444-b88e-a859faea2846","cardType":"basic","points":1,"front":"What defines a surd in terms of its numeric type?","back":"An irrational number expressed as an exact root.","mcqOptions":[{"text":"An irrational number expressed as an exact root.","correct":true,"explanation":"A surd is an exact root whose value is irrational, such as $\\sqrt{2}$. It cannot be written exactly as a fraction of two integers."},{"text":"A rational number expressed as an exact root.","correct":false,"explanation":"A root with a rational value, such as $\\sqrt{9}=3$, is not a surd. A surd is an irrational number expressed as an exact root."},{"text":"An integer number expressed as an exact root.","correct":false,"explanation":"Integers obtained from roots of perfect squares are rational, so they are not surds. A surd must be an irrational number left in exact root form."},{"text":"A decimal number expressed as an exact root.","correct":false,"explanation":"Writing a root as a decimal usually gives an approximation rather than an exact value. A surd is an irrational number expressed exactly using a root symbol."}]},{"id":"15f630ed-c806-4013-a970-554e90f630be","cardType":"basic","points":1,"front":"Why is $\\sqrt{64}$ not considered a surd?","back":"64 is a perfect square, so $\\sqrt{64} = 8$ (a rational number).","mcqOptions":[{"text":"64 is a perfect square, so $\\sqrt{64} = 8$ (a rational number).","correct":true,"explanation":"Since $64=8^2$, its square root is the integer $8$. Surds are irrational roots, so a root that simplifies to a rational number is not a surd."},{"text":"64 is a perfect cube, so $\\sqrt{64} = 4$ (a rational number).","correct":false,"explanation":"Confusing square roots with cube roots gives $4$, since $\\sqrt[3]{64}=4$. In fact, $64=8^2$, so $\\sqrt{64}=8$ and is rational."},{"text":"64 is an even number, so $\\sqrt{64} = 32$ (a rational number).","correct":false,"explanation":"Halving an even number does not calculate its square root. Because $64$ is the perfect square $8^2$, $\\sqrt{64}=8$, which is rational."},{"text":"64 is a multiple of eight, so $\\sqrt{64} = 16$ (a rational number).","correct":false,"explanation":"Being a multiple of eight does not determine a square root, and $16^2$ is $256$. Since $64=8^2$, the correct value is $8$, a rational number."}]},{"id":"04b5e7ac-c1ba-4b74-b6ec-f842aa3c47e8","cardType":"basic","points":1,"front":"State the rule for splitting a product inside a square root.","back":"$$\\sqrt{ab} = \\sqrt{a}\\sqrt{b}$","mcqOptions":[{"text":"$$\\sqrt{ab} = \\sqrt{a}\\sqrt{b}$","correct":true,"explanation":"A product under a square root can be split into the product of the individual square roots, assuming the quantities are non-negative in the IGCSE context."},{"text":"$$\\sqrt{ab} = \\sqrt{a}+\\sqrt{b}$","correct":false,"explanation":"The product rule is often confused with addition, but square roots of factors must be multiplied. The correct identity is $\\sqrt{ab}=\\sqrt{a}\\sqrt{b}$."},{"text":"$$\\sqrt{ab} = a\\sqrt{b}$","correct":false,"explanation":"Taking $a$ outside unchanged incorrectly treats it as a perfect-square factor. The correct general rule is $\\sqrt{ab}=\\sqrt{a}\\sqrt{b}$; only $\\sqrt{a^2b}=a\\sqrt{b}$."},{"text":"$$\\sqrt{ab} = \\sqrt{a+b}$","correct":false,"explanation":"Replacing multiplication inside the root with addition changes the value of the expression. A product splits according to $\\sqrt{ab}=\\sqrt{a}\\sqrt{b}$."}]},{"id":"1113a1e4-b863-4f41-bb9c-7ec2c82d0951","cardType":"basic","points":1,"front":"What is the first step to simplify a surd like $\\sqrt{72}$?","back":"Find the largest perfect-square factor of the number.","mcqOptions":[{"text":"Find the largest perfect-square factor of the number.","correct":true,"explanation":"Using the largest perfect-square factor allows its square root to be taken immediately, leaving the simplest possible surd in one step."},{"text":"Find the largest prime-number factor of the number.","correct":false,"explanation":"A prime factor cannot usually be removed from a square root by itself. The useful factor is the largest perfect-square factor, such as $36$ in $72$."},{"text":"Find the nearest perfect-square number to the number.","correct":false,"explanation":"Using a nearby perfect square would change the original value rather than simplify it exactly. The first step is to find the largest perfect-square factor of the number."},{"text":"Find the decimal approximation of the number.","correct":false,"explanation":"A decimal approximation loses the exact surd form and does not perform algebraic simplification. Instead, identify the largest perfect-square factor inside the root."}]},{"id":"46fda38a-2d26-4ebd-95a1-331b9b4a508f","cardType":"basic","points":2,"front":"Simplify $\\sqrt{72}$ into the form $k\\sqrt{2}$.","back":"$$\\sqrt{36 \\times 2} = 6\\sqrt{2}$","mcqOptions":[{"text":"$$\\sqrt{36 \\times 2} = 6\\sqrt{2}$","correct":true,"explanation":"Since $72=36\\times2$ and $36$ is a perfect square, $\\sqrt{72}=\\sqrt{36}\\sqrt{2}=6\\sqrt{2}$."},{"text":"$$\\sqrt{36 \\times 2} = 3\\sqrt{2}$","correct":false,"explanation":"This incorrectly evaluates $\\sqrt{36}$ as $3$, confusing it with $\\sqrt{9}$. Since $\\sqrt{36}=6$, the correct simplification is $6\\sqrt{2}$."},{"text":"$$\\sqrt{36 \\times 2} = 36\\sqrt{2}$","correct":false,"explanation":"This takes $36$ outside the root without finding its square root. The extracted coefficient must be $\\sqrt{36}=6$, giving $6\\sqrt{2}$."},{"text":"$$\\sqrt{36 \\times 2} = 6\\sqrt{6}$","correct":false,"explanation":"The factor remaining under the root is $2$, not $6$. Splitting $\\sqrt{36\\times2}$ gives $\\sqrt{36}\\sqrt{2}=6\\sqrt{2}$."}]},{"id":"b2e8bef0-dedf-4270-ab74-153f98b2fc01","cardType":"cloze","points":null,"front":"While $\\sqrt{ab} = \\sqrt{a}\\sqrt{b}$ is true, the expression $\\sqrt{a+b}$ [ ] $\\sqrt{a}+\\sqrt{b}$.","back":"While $\\sqrt{ab} = \\sqrt{a}\\sqrt{b}$ is true, the expression $\\sqrt{a+b}$ is not equal to $\\sqrt{a}+\\sqrt{b}$.","mcqOptions":[{"text":"is not equal to","correct":true,"explanation":"Square roots do not distribute over addition. For example, $\\sqrt{9+16}=5$, whereas $\\sqrt{9}+\\sqrt{16}=7$, so the expressions are not equal."},{"text":"is equal to","correct":false,"explanation":"This incorrectly applies the product rule to a sum. Square roots distribute over multiplication, but $\\sqrt{a+b}$ is not equal to $\\sqrt{a}+\\sqrt{b}$."},{"text":"can be expanded to","correct":false,"explanation":"Treating a square root like an algebraic bracket creates a false expansion. The valid product rule does not extend to sums, so the expressions are not equal."},{"text":"is always twice","correct":false,"explanation":"There is no fixed factor of two between these expressions. For instance, $\\sqrt{9+16}=5$ while $\\sqrt{9}+\\sqrt{16}=7$, so they are not equal."}]},{"id":"2108fbe4-2136-4e87-9d94-2568ef35b572","cardType":"basic","points":1,"front":"What are like surds?","back":"Surds that have the same root part (e.g. $3\\sqrt{7}$ and $5\\sqrt{7}$).","mcqOptions":[{"text":"Surds that have the same root part (e.g. $3\\sqrt{7}$ and $5\\sqrt{7}$).","correct":true,"explanation":"Like surds contain the same irrational root part, while their coefficients may differ. This allows their coefficients to be added or subtracted like algebraic terms."},{"text":"Surds that have the same coefficient (e.g. $3\\sqrt{7}$ and $3\\sqrt{5}$).","correct":false,"explanation":"Matching coefficients do not make surds like terms when their root parts differ. Like surds must share the same root part, such as $3\\sqrt{7}$ and $5\\sqrt{7}$."},{"text":"Surds that have different root parts (e.g. $3\\sqrt{7}$ and $5\\sqrt{3}$).","correct":false,"explanation":"Different root parts make the surds unlike, so their coefficients cannot be combined directly. Like surds have the same root part, such as multiples of $\\sqrt{7}$."},{"text":"Surds that have the same value (e.g. $2\\sqrt{3}$ and $\\sqrt{12}$).","correct":false,"explanation":"Equivalent expressions may simplify to the same value, but that is not the definition of like surds. In simplified form, like surds have the same root part."}]},{"id":"6469f1da-b796-4afb-a203-4d7224486837","cardType":"basic","points":1,"front":"Simplify $12\\sqrt{3} + 10\\sqrt{3}$.","back":"$$22\\sqrt{3}$","mcqOptions":[{"text":"$$22\\sqrt{3}$","correct":true,"explanation":"Both terms contain the same root part, $\\sqrt{3}$, so they are like surds. Adding their coefficients gives $(12+10)\\sqrt{3}=22\\sqrt{3}$."},{"text":"$$22\\sqrt{6}$","correct":false,"explanation":"This incorrectly adds the numbers inside identical roots as well as the coefficients. Like surds keep the root part unchanged, giving $(12+10)\\sqrt{3}=22\\sqrt{3}$."},{"text":"$$2\\sqrt{3}$","correct":false,"explanation":"This subtracts the coefficients instead of adding them, producing $12-10=2$. The operation shown is addition, so the correct result is $22\\sqrt{3}$."},{"text":"$$120\\sqrt{3}$","correct":false,"explanation":"This multiplies the coefficients $12$ and $10$ even though the terms are being added. Like surds are combined by adding coefficients, giving $22\\sqrt{3}$."}]},{"id":"3ed523a1-5949-4749-95f5-5cc071247d8a","cardType":"basic","points":2,"front":"How do you simplify a surd with a coefficient, such as $4\\sqrt{45}$?","back":"

Simplify the root: $\\sqrt{45} = 3\\sqrt{5}$
Multiply by the coefficient: $4 \\times 3\\sqrt{5} = 12\\sqrt{5}$

","mcqOptions":[{"text":"

Simplify the root: $\\sqrt{45} = 3\\sqrt{5}$
Multiply by the coefficient: $4 \\times 3\\sqrt{5} = 12\\sqrt{5}$

","correct":true,"explanation":"Since $45=9\\times5$, $\\sqrt{45}=\\sqrt{9}\\sqrt{5}=3\\sqrt{5}$. Multiplying this by the outside coefficient $4$ gives $12\\sqrt{5}$."},{"text":"

Simplify the root: $\\sqrt{45} = 9\\sqrt{5}$
Multiply by the coefficient: $4 \\times 9\\sqrt{5} = 36\\sqrt{5}$

","correct":false,"explanation":"This takes $9$ outside the root without square-rooting it. Since $\\sqrt{9}=3$, $\\sqrt{45}=3\\sqrt{5}$ and therefore $4\\sqrt{45}=12\\sqrt{5}$."},{"text":"

Simplify the root: $\\sqrt{45} = 3\\sqrt{5}$
Add the coefficient: $4 + 3\\sqrt{5} = 7\\sqrt{5}$

","correct":false,"explanation":"The coefficient $4$ multiplies the surd rather than adding to its coefficient. As $\\sqrt{45}=3\\sqrt{5}$, the expression becomes $4\\times3\\sqrt{5}=12\\sqrt{5}$."},{"text":"

Simplify the root: $\\sqrt{45} = 5\\sqrt{3}$
Multiply by the coefficient: $4 \\times 5\\sqrt{3} = 20\\sqrt{3}$

","correct":false,"explanation":"This incorrectly removes $5$ from the root, although $25$ is not a factor of $45$. Using $45=9\\times5$ gives $\\sqrt{45}=3\\sqrt{5}$ and the result $12\\sqrt{5}$."}]},{"id":"6dcdb9a0-c5d4-466f-91f4-47d678079216","cardType":"basic","points":1,"front":"What is the result of $\\sqrt{a} \\times \\sqrt{a}$?","back":"$$a$","mcqOptions":[{"text":"$$a$","correct":true,"explanation":"Multiplying a square root by itself reverses the square-root operation: $\\sqrt{a}\\times\\sqrt{a}=(\\sqrt{a})^2=a$, for values where the square root is defined."},{"text":"$$a^2$","correct":false,"explanation":"This squares $a$ rather than squaring its square root. Since $(\\sqrt{a})^2=a$, the product $\\sqrt{a}\\times\\sqrt{a}$ simplifies to $a$."},{"text":"$$2a$","correct":false,"explanation":"The factor $2$ usually comes from adding two identical terms, not multiplying them. Multiplying $\\sqrt{a}$ by itself gives $(\\sqrt{a})^2=a$."},{"text":"$$\\sqrt{2a}$","correct":false,"explanation":"This treats multiplication like addition inside the root. The correct rule gives $\\sqrt{a}\\times\\sqrt{a}=\\sqrt{a^2}=a$, assuming $a$ is non-negative."}]},{"id":"f16db7ad-d212-426d-ae3a-c27133f0f799","cardType":"cloze","points":null,"front":"Two expressions such as $a+\\sqrt{b}$ and $a-\\sqrt{b}$ are known as [ ].","back":"Two expressions such as $a+\\sqrt{b}$ and $a-\\sqrt{b}$ are known as conjugates.","mcqOptions":[{"text":"conjugates","correct":true,"explanation":"Conjugates contain the same two terms but have opposite signs between them. Thus $a+\\sqrt{b}$ and $a-\\sqrt{b}$ form a conjugate pair."},{"text":"reciprocals","correct":false,"explanation":"Reciprocals multiply to $1$, whereas these expressions differ only in the sign between their terms. They are called conjugates, and their product eliminates the surd cross-terms."},{"text":"factors","correct":false,"explanation":"Factors are expressions multiplied together to form a product, so the term does not describe this sign-changing relationship. The two expressions are known as conjugates."},{"text":"coefficients","correct":false,"explanation":"Coefficients are numerical or algebraic multipliers of terms. Expressions with identical terms and opposite signs, such as $a+\\sqrt{b}$ and $a-\\sqrt{b}$, are conjugates."}]},{"id":"892534ad-0e8c-444f-93d9-6e5ef85309d2","cardType":"basic","points":2,"front":"What is the simplified result of the product $(a+\\sqrt{b})(a-\\sqrt{b})$?","back":"$$a^2 - b$","mcqOptions":[{"text":"$$a^2 - b$","correct":true,"explanation":"The expressions are conjugates, so the middle terms $-a\\sqrt{b}$ and $+a\\sqrt{b}$ cancel. The remaining terms are $a^2-(\\sqrt{b})^2=a^2-b$."},{"text":"$$a^2 + b$","correct":false,"explanation":"The final terms multiply to $\\sqrt{b}\\times(-\\sqrt{b})=-b$, not $+b$. After the middle terms cancel, the correct product is $a^2-b$."},{"text":"$$a^2 - 2b$","correct":false,"explanation":"The two middle terms cancel rather than combining to produce another $b$. Since $(\\sqrt{b})^2=b$, the conjugate product simplifies to $a^2-b$."},{"text":"$$a - b^2$","correct":false,"explanation":"This squares the wrong parts of the product. Multiplying the first terms gives $a^2$, while multiplying the surd terms gives $-b$, producing $a^2-b$."}]},{"id":"529523e8-b7b5-4a67-8d7b-da363f2a9db0","cardType":"basic","points":1,"front":"When squaring a bracket like $(3-\\sqrt{5})^2$, why is it helpful to write it as two identical brackets?","back":"To ensure the middle terms are included in the expansion.","mcqOptions":[{"text":"To ensure the middle terms are included in the expansion.","correct":true,"explanation":"Writing $(3-\\sqrt{5})^2$ as $(3-\\sqrt{5})(3-\\sqrt{5})$ makes both cross-products visible. These middle terms are essential and combine to give $-6\\sqrt{5}$."},{"text":"To ensure the middle terms cancel in the expansion.","correct":false,"explanation":"The middle terms cancel when multiplying conjugates, but these brackets are identical rather than conjugate. Here both cross-products are negative, so they combine to give $-6\\sqrt{5}$."},{"text":"To ensure the surd term is removed before the expansion.","correct":false,"explanation":"The surd should not be removed before expanding, and the final result still contains a surd term. Two brackets help ensure both middle cross-products are included."},{"text":"To ensure the constant term is not squared in the expansion.","correct":false,"explanation":"The constant term must be squared: $3\\times3=9$. Writing two identical brackets is helpful because it displays both middle cross-products, preventing them from being omitted."}]},{"id":"4d8b48b5-7c24-4416-8186-1cb4e5f3b569","cardType":"basic","points":2,"front":"Expand and simplify $(5+\\sqrt{6})(5-\\sqrt{6})$.","back":"$$25 - 6 = 19$","mcqOptions":[{"text":"$$25 - 6 = 19$","correct":true,"explanation":"The brackets are conjugates, so the middle terms cancel. The remaining product is $5^2-(\\sqrt{6})^2=25-6=19$."},{"text":"$$25 + 6 = 31$","correct":false,"explanation":"The surd terms have opposite signs, so their product is $\\sqrt{6}\\times(-\\sqrt{6})=-6$, not $+6$. Therefore the result is $25-6=19$."},{"text":"$$10 - 6 = 4$","correct":false,"explanation":"This doubles $5$ instead of multiplying $5$ by itself. The constant product is $5^2=25$, so the conjugate product is $25-6=19$."},{"text":"$$25 - 36 = -11$","correct":false,"explanation":"This incorrectly squares $6$ rather than squaring $\\sqrt{6}$. Since $(\\sqrt{6})^2=6$, the correct calculation is $25-6=19$."}]},{"id":"b489744a-00eb-4e7f-982a-36eeeec37bac","cardType":"basic","points":1,"front":"What is the objective of rationalising a denominator?","back":"To rewrite the fraction so there is no surd in the denominator.","mcqOptions":[{"text":"To rewrite the fraction so there is no surd in the denominator.","correct":true,"explanation":"Rationalising changes an irrational denominator into a rational one without changing the fraction’s value. The resulting denominator therefore contains no surd."},{"text":"To rewrite the fraction so there is no surd in the numerator.","correct":false,"explanation":"Rationalising concerns the denominator, and the numerator may still contain a surd afterwards. Its objective is to produce an equivalent fraction with no surd in the denominator."},{"text":"To rewrite the fraction so the denominator is equal to 1.","correct":false,"explanation":"Rationalising does not generally make the denominator $1$; it makes the denominator rational. The required equivalent fraction has no surd in its denominator."},{"text":"To rewrite the fraction so there are no fractions in the expression.","correct":false,"explanation":"The expression usually remains a fraction after rationalising. The process specifically removes surds from the denominator while preserving the fraction’s original value."}]},{"id":"90dbe8ce-b99b-4c03-a4c4-0dfd62c96d2f","cardType":"basic","points":1,"front":"How do you rationalise a fraction with a single surd denominator $\\frac{k}{\\sqrt{a}}$?","back":"Multiply both the numerator and denominator by $\\sqrt{a}$.","mcqOptions":[{"text":"Multiply both the numerator and denominator by $\\sqrt{a}$.","correct":true,"explanation":"Multiplying by $\\frac{\\sqrt{a}}{\\sqrt{a}}$ preserves the fraction’s value, while the denominator becomes $\\sqrt{a}\\times\\sqrt{a}=a$, which is rational."},{"text":"Multiply only the numerator by $\\sqrt{a}$.","correct":false,"explanation":"Multiplying only the numerator changes the value of the fraction and leaves the denominator irrational. Both parts must be multiplied by $\\sqrt{a}$, giving denominator $a$."},{"text":"Multiply only the denominator by $\\sqrt{a}$.","correct":false,"explanation":"Although this would make the denominator $a$, changing only the denominator changes the fraction’s value. Multiply both numerator and denominator by $\\sqrt{a}$ instead."},{"text":"Multiply both the numerator and denominator by $a$.","correct":false,"explanation":"Multiplying by $a$ leaves the denominator as $a\\sqrt{a}$, which still contains a surd. Multiplying both parts by $\\sqrt{a}$ makes the denominator equal to $a$."}]},{"id":"a66b647e-3275-488c-a45c-8dca5ce693f1","cardType":"cloze","points":null,"front":"To rationalise a denominator with two terms, such as $3+\\sqrt{2}$, you must multiply the numerator and denominator by the [ ].","back":"To rationalise a denominator with two terms, such as $3+\\sqrt{2}$, you must multiply the numerator and denominator by the conjugate.","mcqOptions":[{"text":"conjugate","correct":true,"explanation":"The conjugate of $3+sqrt{2}$ is $3-sqrt{2}$. Multiplying them produces $9-2=7$, eliminating the surd from the denominator."},{"text":"reciprocal","correct":false,"explanation":"A reciprocal swaps the numerator and denominator but does not necessarily remove a surd. The correct multiplier is the conjugate, which uses the opposite sign between the two terms."},{"text":"denominator","correct":false,"explanation":"Multiplying by the denominator itself gives $(3+sqrt{2})^2$, which still contains a surd term. The conjugate $3-sqrt{2}$ creates a difference of two squares."},{"text":"numerator","correct":false,"explanation":"The numerator is unrelated to the method for eliminating a two-term surd denominator. Multiplying by the conjugate makes the denominator rational through the difference-of-squares identity."}]},{"id":"fb8cc97e-88ba-4713-8dcd-a3865f54f87c","cardType":"basic","points":2,"front":"Simplify $\\frac{12}{\\sqrt{3}}$ fully.","back":"$$\\frac{12\\sqrt{3}}{3} = 4\\sqrt{3}$","mcqOptions":[{"text":"$$\\frac{12\\sqrt{3}}{3} = 4\\sqrt{3}$","correct":true,"explanation":"Multiplying top and bottom by $sqrt{3}$ gives $frac{12sqrt{3}}{3}$. Dividing the coefficient $12$ by $3$ then produces the fully simplified result $4sqrt{3}$."},{"text":"$$frac{12sqrt{3}}{3} = 12sqrt{3}$","correct":false,"explanation":"The denominator becomes $3$, so the coefficient must also be divided by $3$. Failing to perform this division gives $12sqrt{3}$ instead of the correct $4sqrt{3}$."},{"text":"$$frac{12sqrt{3}}{6} = 2sqrt{3}$","correct":false,"explanation":"Multiplying $sqrt{3}$ by itself gives $3$, not $6$. Rationalising therefore gives $frac{12sqrt{3}}{3}=4sqrt{3}$, rather than halving the correct coefficient."},{"text":"$$frac{12}{3} = 4$","correct":false,"explanation":"This treats $sqrt{3}$ as though it were $3$. Rationalising requires multiplying by $sqrt{3}$, giving $frac{12sqrt{3}}{3}=4sqrt{3}$."}]},{"id":"c5835a17-f75d-43bb-84c0-41219914d5de","cardType":"basic","points":2,"front":"Simplify the algebraic surd expression $(\\sqrt{x}+\\sqrt{y})(\\sqrt{x}-\\sqrt{y})$.","back":"$$x - y$","mcqOptions":[{"text":"$$x - y$","correct":true,"explanation":"The expression is a product of conjugates, so it follows $(a+b)(a-b)=a^2-b^2$. Thus $(sqrt{x})^2-(sqrt{y})^2=x-y$."},{"text":"$$x + y$","correct":false,"explanation":"The signs of the two cross terms are opposite, so they cancel; the final term is negative. Applying the difference-of-squares identity gives $x-y$, not $x+y$."},{"text":"$$x - 2sqrt{xy} + y$","correct":false,"explanation":"This is the expansion of $(sqrt{x}-sqrt{y})^2$, not a product of conjugates. Here the cross terms cancel completely, leaving the correct result $x-y$."},{"text":"$$sqrt{x} - sqrt{y}$","correct":false,"explanation":"This leaves the factors essentially unexpanded and ignores that each square root is multiplied by itself. Full expansion cancels the cross terms and gives $x-y$."}]},{"id":"78646f5d-cece-476d-83e3-e5360a1fa023","cardType":"basic","points":2,"front":"Expand and simplify $(3p+\\sqrt{q})^2$.","back":"$$9p^2 + 6p\\sqrt{q} + q$","mcqOptions":[{"text":"$$9p^2 + 6p\\sqrt{q} + q$","correct":true,"explanation":"Using $(a+b)^2=a^2+2ab+b^2$, the terms are $(3p)^2=9p^2$, $2(3p)(sqrt{q})=6psqrt{q}$ and $(sqrt{q})^2=q$."},{"text":"$$9p^2 + q$","correct":false,"explanation":"Squaring a binomial requires the cross term $2ab$. Here it is $2(3p)(sqrt{q})=6psqrt{q}$, so the complete result is $9p^2+6psqrt{q}+q$."},{"text":"$$9p^2 + 3psqrt{q} + q$","correct":false,"explanation":"The middle term must include a factor of $2$ because both cross-products occur. Therefore it is $6psqrt{q}$, giving $9p^2+6psqrt{q}+q$."},{"text":"$$9p^2 + 6psqrt{q} + q^2$","correct":false,"explanation":"Squaring $sqrt{q}$ gives $q$, not $q^2$. The first and middle terms are correct, but the full expansion ends with $+q$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Surds","title":"Surds Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"93bf8b6f-a42a-4df9-a766-7beba6eb1d7f","walled":"$L39"}],"$L3a","$L3b"]