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with $a \\neq 0$."},{"text":"$$ax^2+bx+c=x$","correct":false,"explanation":"Leaving an $x$ on the right does not give standard form. All terms must be collected on one side to produce $ax^2+bx+c=0$."},{"text":"$$ax^2+bx=c$","correct":false,"explanation":"This leaves the constant on the right-hand side rather than collecting every term on one side. Standard form is $ax^2+bx+c=0$."},{"text":"$$ax^3+bx+c=0$","correct":false,"explanation":"Using $x^3$ makes the equation cubic, not quadratic. A quadratic has highest power $2$, so its standard form is $ax^2+bx+c=0$."}]},{"id":"49c43d08-40be-4cb9-b3d4-a86195810bf1","cardType":"basic","points":1,"front":"What is the first step required before using the quadratic formula?","back":"Rearrange the equation into standard form ($ax^2+bx+c=0$).","mcqOptions":[{"text":"Rearrange the equation into standard form ($ax^2+bx+c=0$).","correct":true,"explanation":"The coefficients $a$, $b$, and $c$ can only be identified reliably after every term is collected on one side and the equation equals zero."},{"text":"Rearrange the equation into factorised form ($(ax+b)(cx+d)=0$).","correct":false,"explanation":"Factorised form is not required and may be difficult or impossible to obtain neatly. The quadratic formula requires standard form, $ax^2+bx+c=0$."},{"text":"Rearrange the equation into completed square form ($a(x+b)^2+c=0$).","correct":false,"explanation":"Completing the square is a different solving method and is unnecessary before applying the formula. First rearrange the equation into $ax^2+bx+c=0$."},{"text":"Rearrange the equation into linear form ($ax+b=0$).","correct":false,"explanation":"Removing the squared term changes the equation and loses its quadratic solutions. Instead, retain all terms and rearrange into standard form, $ax^2+bx+c=0$."}]},{"id":"d7ea47c7-2c4f-2106-cb69-b9bea9ebdd99","cardType":"basic","points":1,"front":"Why is the quadratic formula used instead of factorising?","back":"It works for any quadratic, including those with decimal or surd answers that do not factorise neatly.","mcqOptions":[{"text":"It works for any quadratic, including those with decimal or surd answers that do not factorise neatly.","correct":true,"explanation":"The quadratic formula applies to every quadratic written in standard form, so it remains effective when integer factorisation is not available and roots are irrational or decimal."},{"text":"It works only for factorisable quadratics, including those with integer answers that can be found using brackets neatly.","correct":false,"explanation":"Restricting the formula to factorisable quadratics reverses its main advantage. It works for any quadratic, including equations whose decimal or surd roots prevent neat factorisation."},{"text":"It works for linear equations, including those with decimal or fraction answers that do not rearrange neatly.","correct":false,"explanation":"The quadratic formula is specifically derived for equations containing an $x^2$ term, not linear equations. It solves any quadratic, including ones that do not factorise neatly."},{"text":"It works for some quadratics, including only those with whole-number answers that do not require square roots.","correct":false,"explanation":"The formula is not limited to whole-number roots. It works for any quadratic and can produce fractions, decimals, surds, or no real solutions."}]},{"id":"55ce26e5-4003-7151-51f7-952cdb69ce6e","cardType":"basic","points":1,"front":"State the quadratic formula used to solve $ax^2+bx+c=0$.","back":"$$x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$","mcqOptions":[{"text":"$$x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$","correct":true,"explanation":"The formula gives both roots by using plus and minus before the square root of the discriminant, with the entire numerator divided by $2a$."},{"text":"$$x=\\frac{b\\pm\\sqrt{b^2-4ac}}{2a}$","correct":false,"explanation":"A common sign error is using $b$ instead of its additive inverse. The numerator must begin with $-b$, giving $x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$."},{"text":"$$x=\\frac{-b\\pm\\sqrt{b^2+4ac}}{2a}$","correct":false,"explanation":"The discriminant uses subtraction, not addition. The correct expression under the square root is $b^2-4ac$, so the formula is $x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$."},{"text":"$$x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2}$","correct":false,"explanation":"Omitting $a$ from the denominator incorrectly changes both roots unless $a=1$. The whole numerator must be divided by $2a$."}]},{"id":"9879bceb-fefb-a0a7-8821-94a6fa0285b5","cardType":"cloze","points":null,"front":"In the quadratic formula, the expression $\\text{[...]}$ is known as the discriminant.","back":"In the quadratic formula, the expression $\\boxed{b^2-4ac}$ is known as the discriminant.","mcqOptions":[{"text":"$$b^2-4ac$","correct":true,"explanation":"The discriminant is the expression beneath the square root in the quadratic formula. Its value determines the number and type of solutions to the quadratic."},{"text":"$$b^2+4ac$","correct":false,"explanation":"Confusing subtraction with addition changes the value under the square root. The discriminant is $b^2-4ac$, which determines the number and type of roots."},{"text":"$$b-4ac$","correct":false,"explanation":"The coefficient $b$ must be squared in the discriminant. Omitting this square is a formula-recall error; the correct expression is $b^2-4ac$."},{"text":"$$b^2-2ac$","correct":false,"explanation":"Using $2ac$ confuses the denominator $2a$ with the discriminant. The expression under the square root is $b^2-4ac$."}]},{"id":"e9be3418-3829-f558-f163-9fef0338968c","cardType":"basic","points":2,"front":"What are the coefficients $a$, $b$, and $c$ for the equation $5x^2=4x+7$?","back":"

$a=5$
$b=-4$
$c=-7$

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

$a=5$
$b=-4$
$c=-7$

","correct":true,"explanation":"Rearranging gives $5x^2-4x-7=0$. Comparing this with $ax^2+bx+c=0$ shows that $a=5$, $b=-4$, and $c=-7$."},{"text":"

$a=5$
$b=4$
$c=7$

","correct":false,"explanation":"These signs are taken before moving the right-hand terms. Rearranging gives $5x^2-4x-7=0$, so $b=-4$ and $c=-7$."},{"text":"

$a=-5$
$b=4$
$c=7$

","correct":false,"explanation":"Moving $5x^2$ to the right can produce these coefficients, but the equation would then need consistent standard form. Using $5x^2-4x-7=0$ gives $a=5$, $b=-4$, $c=-7$."},{"text":"

$a=5$
$b=-4$
$c=7$

","correct":false,"explanation":"Only changing the sign of the $4x$ term overlooks that both right-hand terms move left. Rearranging gives $5x^2-4x-7=0$, so $c=-7$."}]},{"id":"e8c9bf47-e931-490a-cf03-90c8bffa6ee6","cardType":"cloze","points":null,"front":"The symbol $\\pm$ in the quadratic formula means [ ], which usually results in two solutions.","back":"The symbol $\\pm$ in the quadratic formula means plus or minus, which usually results in two solutions.","mcqOptions":[{"text":"plus or minus","correct":true,"explanation":"The symbol instructs you to calculate once using addition and once using subtraction. These two calculations usually produce the two solutions of a quadratic equation."},{"text":"plus then multiply","correct":false,"explanation":"The second symbol is a minus sign, not an instruction to multiply. The notation means plus or minus, requiring separate calculations for the two possible roots."},{"text":"positive or negative","correct":false,"explanation":"This confuses the operation with the signs of the resulting answers, which need not be one positive and one negative. The symbol means plus or minus."},{"text":"plus and minus","correct":false,"explanation":"Treating both operations as simultaneous is misleading. The notation means plus or minus: calculate one solution with addition and another with subtraction."}]},{"id":"bfd69a73-ca5f-00f6-0169-5eccafeb14d4","cardType":"basic","points":1,"front":"Why should you write out the substitution step in an exam before using a calculator?","back":"To ensure you get method marks even if the final calculation is wrong.","mcqOptions":[{"text":"To ensure you get method marks even if the final calculation is wrong.","correct":true,"explanation":"A correct substitution demonstrates that the formula and coefficients were used properly. Examiners can therefore award method marks even when a later calculator or rounding error occurs."},{"text":"To ensure you get accuracy marks even if the formula you selected is wrong.","correct":false,"explanation":"Accuracy marks require a correct result or valid stage, not merely written substitution into a wrong formula. Showing correct substitution protects method marks if later calculation goes wrong."},{"text":"To ensure you get full marks even if the coefficients you substituted are wrong.","correct":false,"explanation":"Writing an incorrect substitution cannot guarantee full marks because wrong coefficients undermine the method. Correctly showing the substitution can secure method marks despite a later calculation error."},{"text":"To ensure you get presentation marks even if no final solutions are written down.","correct":false,"explanation":"GCSE quadratic questions generally reward mathematical method rather than presentation alone. Writing the substitution can earn method marks, but the solutions should still be calculated and stated."}]},{"id":"328b66b8-c5a4-8b80-bf9b-1ebbd59c11bb","cardType":"basic","points":1,"front":"When solving a quadratic, at what stage should you round your answers?","back":"Only at the final step (do not round the square root halfway through).","mcqOptions":[{"text":"Only at the final step (do not round the square root halfway through).","correct":true,"explanation":"Keeping the square root and intermediate values exact prevents rounding errors from accumulating. Round only the final solutions to the accuracy requested."},{"text":"At the square root step (round the square root before continuing).","correct":false,"explanation":"Rounding the square root early introduces an approximation that can affect later calculations. Keep full calculator precision and round only at the final step."},{"text":"After calculating each term (round every value before combining them).","correct":false,"explanation":"Rounding each term causes several small errors to accumulate in the final solutions. Intermediate values should remain exact or unrounded until the final step."},{"text":"Before using the quadratic formula (round all the coefficients first).","correct":false,"explanation":"Rounding the coefficients changes the original quadratic and may produce inaccurate roots. Use the given coefficients and round only the final answers."}]},{"id":"3473cc4a-edc2-99ad-73d2-2c7daa34bf6a","cardType":"basic","points":1,"front":"What is a surd?","back":"An exact square root that cannot be simplified to a whole number.","mcqOptions":[{"text":"An exact square root that cannot be simplified to a whole number.","correct":true,"explanation":"A surd represents an irrational root exactly rather than as a rounded decimal, such as $\\sqrt{3}$. It cannot be simplified to a whole number."},{"text":"An approximate square root that has been rounded to a decimal number.","correct":false,"explanation":"A rounded decimal is only an approximation, whereas a surd preserves the exact value. A surd is an exact square root that cannot simplify to a whole number."},{"text":"An exact square root that can be simplified to a whole number.","correct":false,"explanation":"Roots such as $\\sqrt{25}=5$ simplify to whole numbers and are therefore not surds. A surd cannot be simplified to a whole number."},{"text":"An exact cube root that can always be written as a whole number.","correct":false,"explanation":"This confuses square roots with cube roots and incorrectly requires a whole-number result. A surd is an exact root that cannot simplify to a rational number."}]},{"id":"d0f5bcb5-64f4-5c3e-29dd-a4684ebf8c0a","cardType":"basic","points":2,"front":"How do you simplify a surd like $\\sqrt{108}$?","back":"Find the largest square number that divides into it (e.g., $\\sqrt{36 \\times 3} = 6\\sqrt{3}$).","mcqOptions":[{"text":"Find the largest square number that divides into it (e.g., $\\sqrt{36 \\times 3} = 6\\sqrt{3}$).","correct":true,"explanation":"Factoring out the largest square factor simplifies the surd fully. Since $108=36\\times3$, $\\sqrt{108}=\\sqrt{36}\\sqrt{3}=6\\sqrt{3}$."},{"text":"Find the smallest square number that divides into it (e.g., $\\sqrt{4 \\times 27} = 2\\sqrt{27}$).","correct":false,"explanation":"Using a smaller square factor may leave another square factor inside the root, so the result is not fully simplified. The largest square factor of $108$ is $36$."},{"text":"Find the largest prime number that divides into it (e.g., $\\sqrt{3 \\times 36} = 3\\sqrt{36}$).","correct":false,"explanation":"Only square factors can be taken out of a square root directly; a factor of $3$ does not become $3$. Using $36$ gives $\\sqrt{108}=6\\sqrt{3}$."},{"text":"Find the nearest square number to its value (e.g., $\\sqrt{100 + 8} = 10\\sqrt{8}$).","correct":false,"explanation":"Square roots do not distribute over addition, so $\\sqrt{100+8}$ is not $10\\sqrt{8}$. Factorise using the largest square divisor: $108=36\\times3$."}]},{"id":"16c6222d-eaa3-f212-6613-c47ed71960dd","cardType":"basic","points":1,"front":"How do you give an exact answer for a quadratic equation?","back":"Leave the answer in surd form rather than converting to a decimal.","mcqOptions":[{"text":"Leave the answer in surd form rather than converting to a decimal.","correct":true,"explanation":"Surd form preserves the exact value of an irrational root. Converting it to a terminating calculator decimal would round the value and make it approximate."},{"text":"Give the answer in decimal form rounded to three significant figures.","correct":false,"explanation":"A rounded decimal is an approximation, even when several significant figures are included. An exact quadratic solution containing an irrational root must remain in surd form."},{"text":"Leave the answer in fraction form after removing every square root.","correct":false,"explanation":"An irrational solution cannot be represented exactly as an ordinary fraction, so removing every square root is impossible. The exact answer should be left in surd form."},{"text":"Give the answer in integer form by rounding to the nearest whole number.","correct":false,"explanation":"Rounding an irrational solution to an integer loses accuracy and does not give an exact value. Exact quadratic solutions should remain in surd form."}]},{"id":"7f916cba-6ead-b571-4219-af7042be1ab5","cardType":"basic","points":2,"front":"When solving for a length in a geometry problem, why might you reject one of the quadratic roots?","back":"Lengths cannot be negative; any root that results in a negative side length is invalid in context.","mcqOptions":[{"text":"Lengths cannot be negative; any root that results in a negative side length is invalid in context.","correct":true,"explanation":"A quadratic may produce two algebraic roots, but a physical length must be non-negative. Any root giving a negative side length is rejected in the geometry context."},{"text":"Lengths must be whole numbers; any root that results in a decimal side length is invalid in context.","correct":false,"explanation":"Lengths may be decimals, fractions or surds, so a non-integer root can still be valid. Only a root producing a negative length must be rejected."},{"text":"Lengths cannot be irrational; any root that results in a surd side length is invalid in context.","correct":false,"explanation":"Geometric lengths can be irrational, as diagonals often demonstrate, so a surd length may be exact and valid. A root is rejected if it gives a negative length."},{"text":"Lengths must use both roots; any solution that rejects one side length is invalid in context.","correct":false,"explanation":"Although a quadratic can have two roots, both need not satisfy the physical conditions of the problem. A root producing a negative length must be rejected."}]},{"id":"4abac2b9-6195-520d-3ab0-4394a16da8e9","cardType":"basic","points":2,"front":"How is Pythagoras' Theorem used to form a quadratic?","back":"Substitute side expressions into $a^2+b^2=c^2$ and expand/rearrange into $ax^2+bx+c=0$.","mcqOptions":[{"text":"Substitute side expressions into $a^2+b^2=c^2$ and expand/rearrange into $ax^2+bx+c=0$.","correct":true,"explanation":"Pythagoras’ Theorem relates the squared side lengths of a right-angled triangle. Substituting algebraic expressions, expanding and rearranging can therefore produce a quadratic equation."},{"text":"Substitute side expressions into $a+b=c$ and expand/rearrange into $ax^2+bx+c=0$.","correct":false,"explanation":"This omits the squares required by Pythagoras’ Theorem. For a right-angled triangle, substitute the side expressions into $a^2+b^2=c^2$, then expand and rearrange."},{"text":"Substitute side expressions into $a^2-b^2=c^2$ and expand/rearrange into $ax^2+bx+c=0$.","correct":false,"explanation":"This uses subtraction instead of the Pythagorean sum. The correct relationship is $a^2+b^2=c^2$, with $c$ representing the hypotenuse."},{"text":"Substitute side expressions into $2a+2b=c$ and expand/rearrange into $ax^2+bx+c=0$.","correct":false,"explanation":"This resembles part of a perimeter formula rather than Pythagoras’ Theorem. Use $a^2+b^2=c^2$ for a right-angled triangle, then expand and rearrange."}]},{"id":"d312cfa3-d109-c5d4-e915-fc4343aeb73b","cardType":"cloze","points":null,"front":"To find significant figures, you must start counting from the [ ].","back":"To find significant figures, you must start counting from the first non-zero digit.","mcqOptions":[{"text":"first non-zero digit","correct":true,"explanation":"Leading zeros only locate the decimal point and are not significant. Counting begins with the first digit that is not zero, wherever it appears."},{"text":"first digit after the decimal point","correct":false,"explanation":"Decimal places begin immediately after the decimal point, but significant figures may begin before or several places after it. Start at the first non-zero digit."},{"text":"first zero digit","correct":false,"explanation":"Leading zeros are placeholders and do not count as significant figures. Counting starts from the first non-zero digit, after any leading zeros."},{"text":"last non-zero digit","correct":false,"explanation":"Starting at the last non-zero digit would omit earlier significant digits and reverse the counting process. Significant figures are counted from the first non-zero digit."}]},{"id":"968e6ec5-bd8a-a434-35ee-9ae959e07ee3","cardType":"basic","points":1,"front":"If $x=\\frac{-10\\pm6\\sqrt{3}}{2}$, what is the simplified exact form?","back":"$$x=-5\\pm3\\sqrt{3}$","mcqOptions":[{"text":"$$x=-5\\pm3\\sqrt{3}$","correct":true,"explanation":"Both terms in the numerator are divided by $2$: $-10\\div2=-5$ and $6\\sqrt{3}\\div2=3\\sqrt{3}$, giving the two exact solutions."},{"text":"$$x=-5\\pm6\\sqrt{3}$","correct":false,"explanation":"This divides $-10$ by $2$ but forgets to divide the surd term. Dividing both numerator terms gives $x=-5\\pm3\\sqrt{3}$."},{"text":"$$x=-10\\pm3\\sqrt{3}$","correct":false,"explanation":"This divides the surd term by $2$ but leaves the first term unchanged. The denominator divides both terms, giving $x=-5\\pm3\\sqrt{3}$."},{"text":"$$x=5\\pm3\\sqrt{3}$","correct":false,"explanation":"This loses the negative sign when dividing $-10$ by $2$. Since $-10\\div2=-5$, the simplified exact form is $x=-5\\pm3\\sqrt{3}$."}]},{"id":"bc133c08-78f8-b7af-91d6-6c43114fe402","cardType":"basic","points":1,"front":"What is the coefficient $b$ in the equation $x^2-x-9=0$?","back":"$$-1$","mcqOptions":[{"text":"$$-1$","correct":true,"explanation":"In standard form $ax^2+bx+c=0$, $b$ is the coefficient multiplying $x$. Since the middle term is $-x=-1x$, the coefficient is $-1$."},{"text":"$$1$","correct":false,"explanation":"This ignores the negative sign before $x$. Writing the term as $-1x$ shows that the coefficient $b$ is $-1$, not $1$."},{"text":"$$-9$","correct":false,"explanation":"This confuses the constant term $c$ with the coefficient $b$. In $x^2-x-9=0$, the coefficient of $x$ is $-1$, while $c=-9$."},{"text":"$$0$","correct":false,"explanation":"This incorrectly treats the coefficient as missing. The term $-x$ means $-1x$, so the coefficient $b$ is $-1$, not $0$."}]},{"id":"67800988-8807-912b-ef94-762bedc52d09","cardType":"basic","points":1,"front":"In a shape problem, if a rectangle has area 40, width $(x-2)$, and height $(2x+3)$, what is the initial equation?","back":"$$(x-2)(2x+3)=40$","mcqOptions":[{"text":"$$(x-2)(2x+3)=40$","correct":true,"explanation":"A rectangle’s area equals width multiplied by height. Substituting the given expressions and area therefore gives $(x-2)(2x+3)=40$."},{"text":"$$(x-2)+(2x+3)=40$","correct":false,"explanation":"This adds the side lengths rather than multiplying them, confusing an area calculation with part of a perimeter calculation. Area is width times height, so $(x-2)(2x+3)=40$."},{"text":"$$2(x-2)+2(2x+3)=40$","correct":false,"explanation":"This forms the rectangle’s perimeter by doubling and adding its side lengths. The given value is the area, so the required equation is $(x-2)(2x+3)=40$."},{"text":"$$(x+2)(2x-3)=40$","correct":false,"explanation":"This reverses both signs in the given dimensions. The width and height must be copied exactly before multiplying, giving $(x-2)(2x+3)=40$."}]},{"id":"7896ed61-31a9-675b-67e1-50a0b5fd4e9d","cardType":"basic","points":2,"front":"What is the difference between rounding to 3 decimal places (DP) and 3 significant figures (SF)?","back":"

3 DP: counts 3 digits after the decimal point.
3 SF: counts 3 digits starting from the first non-zero number.

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

3 DP: counts 3 digits after the decimal point.
3 SF: counts 3 digits starting from the first non-zero number.

","correct":true,"explanation":"Decimal places are counted from the decimal point, whereas significant figures begin at the first non-zero digit. Zeros after that first significant digit may also count."},{"text":"

3 DP: counts 3 digits starting from the first non-zero number.
3 SF: counts 3 digits after the decimal point.

","correct":false,"explanation":"This swaps the two definitions. Decimal places count digits after the decimal point, while significant figures count from the first non-zero digit."},{"text":"

3 DP: counts 3 digits before the decimal point.
3 SF: counts 3 digits starting from the first non-zero number.

","correct":false,"explanation":"The significant-figures definition is right, but decimal places are not digits before the decimal point. They are counted to the right of the decimal point."},{"text":"

3 DP: counts 3 non-zero digits after the decimal point.
3 SF: counts 3 digits starting from the decimal point.

","correct":false,"explanation":"Decimal places include zero digits and count positions after the decimal point. Significant figures instead begin at the first non-zero digit, wherever it occurs."}]},{"id":"568a6768-e438-2db9-fb87-e52db894fc20","cardType":"cloze","points":null,"front":"In Pythagoras problems, the side used as $c$ in $a^2+b^2=c^2$ must always be the [ ].","back":"In Pythagoras problems, the side used as $c$ in $a^2+b^2=c^2$ must always be the hypotenuse.","mcqOptions":[{"text":"hypotenuse","correct":true,"explanation":"The hypotenuse is the longest side of a right-angled triangle and lies opposite the right angle. It is always represented by $c$ in $a^2+b^2=c^2$."},{"text":"shortest side","correct":false,"explanation":"The shortest side is one of the two perpendicular sides, not necessarily $c$. In Pythagoras’ theorem, $c$ must be the hypotenuse opposite the right angle."},{"text":"adjacent side","correct":false,"explanation":"The adjacent side depends on which acute angle is being considered and is terminology commonly used in trigonometry. In Pythagoras’ theorem, $c$ is always the hypotenuse."},{"text":"opposite side","correct":false,"explanation":"The opposite side depends on the chosen acute angle, so it is not automatically $c$. The side opposite the right angle is the hypotenuse and must be used as $c$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Component 1"},"heading":"Quadratic Formula","title":"Quadratic Formula Flashcards","topbar":{"backHref":"/gcse/maths/eduqas/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"6f130726-38fd-4b41-97c2-fb904d7bde9d","walled":"$L38"}],"$L39","$L3a"]