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corresponding quadratic equation with the expression set equal to zero."},{"text":"one","correct":false,"explanation":"Setting the expression equal to one does not generally identify where its sign may change. The boundary values are found by setting the quadratic expression equal to zero."},{"text":"infinity","correct":false,"explanation":"Setting the expression equal to infinity cannot produce finite boundary values. Boundaries are the roots of the quadratic, found by setting its expression equal to zero."},{"text":"the discriminant","correct":false,"explanation":"A quadratic expression is not set equal to its discriminant to locate boundaries. Instead, set the expression equal to zero and solve for its roots."}]},{"id":"4f8a905e-c0a1-4539-94e3-b22086b9f407","cardType":"basic","points":1,"front":"What do boundary values do to the number line?","back":"They split it into regions.","mcqOptions":[{"text":"They split it into regions.","correct":true,"explanation":"Each boundary value divides the number line at a possible change of sign, creating separate regions that can be tested to determine where the inequality holds."},{"text":"They mark every solution.","correct":false,"explanation":"Boundary values are not automatically all solutions; strict inequalities exclude them. Their role is to split the number line into regions whose signs can be tested."},{"text":"They remove the negative regions.","correct":false,"explanation":"Boundary values do not remove sections of the number line. They split it into regions, allowing one test value to determine the quadratic’s sign in each region."},{"text":"They identify the positive region.","correct":false,"explanation":"Boundary values alone do not show which region satisfies the inequality. They split the number line into regions, after which signs or test values identify the solutions."}]},{"id":"9a10e995-a4ff-471f-ada2-101b89d78191","cardType":"basic","points":2,"front":"Find the boundary values for $x^2+7x+12\\gt 0$.","back":"$x=-4$ and $x=-3$.","mcqOptions":[{"text":"$x=-4$ and $x=-3$.","correct":true,"explanation":"Setting the quadratic equal to zero gives $x^2+7x+12=(x+4)(x+3)=0$, so the two boundary values are $x=-4$ and $x=-3$."},{"text":"$x=4$ and $x=3$.","correct":false,"explanation":"These signs are reversed: $(x+4)(x+3)=0$ requires each bracket to equal zero. Therefore, the boundary values are $x=-4$ and $x=-3$."},{"text":"$x=-4$ and $x=3$.","correct":false,"explanation":"This mixes the signs when reading roots from the factors. Since the factorisation is $(x+4)(x+3)$, both roots are negative: $x=-4$ and $x=-3$."},{"text":"$x=-6$ and $x=-2$.","correct":false,"explanation":"These numbers multiply to $12$ but add to $8$, not $7$. The correct factor pair is $4$ and $3$, giving roots $x=-4$ and $x=-3$."}]},{"id":"a084c408-e19f-4f19-b97a-d11924d571f0","cardType":"cloze","points":null,"front":"If the coefficient of $x^2$ is positive, the parabola opens [ ].","back":"If the coefficient of $x^2$ is positive, the parabola opens upwards.","mcqOptions":[{"text":"upwards","correct":true,"explanation":"A positive coefficient of $x^2$ makes the quadratic increase towards positive infinity at both ends, so its graph is a parabola that opens upwards."},{"text":"downwards","correct":false,"explanation":"A parabola opens downwards only when the coefficient of $x^2$ is negative. A positive coefficient means both arms rise, so it opens upwards."},{"text":"sideways","correct":false,"explanation":"A vertical parabola does not open sideways; that shape would arise from expressing $x$ in terms of $y^2$. A positive $x^2$ coefficient makes it open upwards."},{"text":"inwards","correct":false,"explanation":"A parabola cannot open inwards because its two arms separate as they extend. With a positive coefficient of $x^2$, both arms rise and it opens upwards."}]},{"id":"43eb8ef6-829e-4c46-aff6-628bc28cbf29","cardType":"basic","points":1,"front":"For an upward-opening quadratic, where is it positive relative to its roots?","back":"Outside the roots.","mcqOptions":[{"text":"Outside the roots.","correct":true,"explanation":"An upward-opening parabola lies above the $x$-axis to the left of the smaller root and to the right of the larger root, so it is positive outside."},{"text":"Between the roots.","correct":false,"explanation":"Between two distinct roots, an upward-opening parabola lies below the $x$-axis and is negative. It is positive on the two regions outside the roots."},{"text":"At the roots.","correct":false,"explanation":"At the roots, the quadratic equals zero rather than being positive. For an upward-opening quadratic, positive values occur in the regions outside the roots."},{"text":"To the right of the roots.","correct":false,"explanation":"Positivity is not confined to the right of both roots. An upward-opening parabola is positive on both outer regions: left of the smaller root and right of the larger."}]},{"id":"9c3b6faf-f248-4337-98c7-25c58a74f673","cardType":"basic","points":1,"front":"For an upward-opening quadratic, where is it negative relative to its roots?","back":"Between the roots.","mcqOptions":[{"text":"Between the roots.","correct":true,"explanation":"An upward-opening parabola falls below the $x$-axis after crossing the smaller root and rises above it after the larger root, so it is negative between them."},{"text":"Outside the roots.","correct":false,"explanation":"Outside two distinct roots, an upward-opening parabola lies above the $x$-axis and is positive. Its negative values occur in the region between the roots."},{"text":"At the roots.","correct":false,"explanation":"At either root, the quadratic’s value is zero, not negative. For an upward-opening quadratic with two roots, negative values occur strictly between those roots."},{"text":"Near the roots.","correct":false,"explanation":"The quadratic is negative across the whole interval between its roots, not only near one of them. An upward-opening parabola lies below the axis between both roots."}]},{"id":"70658efe-4589-4db8-a98d-01f28ef9ec84","cardType":"cloze","points":null,"front":"$$\\lt$ and $\\gt$ are [ ] inequalities, so boundary values are not included.","back":"$$\\lt$ and $\\gt$ are strict inequalities, so boundary values are not included.","mcqOptions":[{"text":"strict","correct":true,"explanation":"The symbols $\nlt$ and $\ngt$ require values to be genuinely smaller or larger, not equal. They are therefore strict inequalities and exclude boundary values."},{"text":"inclusive","correct":false,"explanation":"Inclusive inequalities allow equality and use $\nle$ or $\nge$. The symbols $\nlt$ and $\ngt$ do not permit equality, so they are described as strict inequalities."},{"text":"non-strict","correct":false,"explanation":"Non-strict inequalities include equality and are written with $\nle$ or $\nge$. Because $\nlt$ and $\ngt$ exclude equality, the correct description is strict."},{"text":"compound","correct":false,"explanation":"A compound inequality combines two inequality conditions; it does not describe whether endpoints are included. The symbols $\nlt$ and $\ngt$ are strict inequalities."}]},{"id":"8facb816-71e6-497a-aa44-29e850858bb5","cardType":"cloze","points":null,"front":"$$\\le$ and $\\ge$ are [ ] inequalities, so boundary values are included.","back":"$$\\le$ and $\\ge$ are inclusive inequalities, so boundary values are included.","mcqOptions":[{"text":"inclusive","correct":true,"explanation":"The symbols $\nle$ and $\nge$ allow equality as well as inequality. Boundary values satisfy the equality case, so these are inclusive inequalities and include them."},{"text":"strict","correct":false,"explanation":"Strict inequalities use $\nlt$ or $\ngt$ and exclude equality. Since $\nle$ and $\nge$ permit equality at boundary values, they are inclusive inequalities."},{"text":"exclusive","correct":false,"explanation":"Exclusive would mean the boundary values are left out, as with $\nlt$ and $\ngt$. The symbols $\nle$ and $\nge$ include equality, so they are inclusive."},{"text":"open","correct":false,"explanation":"Open usually describes intervals whose endpoints are excluded. Because $\nle$ and $\nge$ include equality and therefore include boundary values, these inequalities are inclusive."}]},{"id":"71216d4e-d094-4225-8bfe-720cd7491990","cardType":"basic","points":1,"front":"On a number line, what endpoint is used for $\\lt$ or $\\gt$?","back":"An open circle.","mcqOptions":[{"text":"An open circle.","correct":true,"explanation":"A strict inequality excludes the endpoint itself, so an open circle is used to show that the boundary value is not part of the solution."},{"text":"A closed circle.","correct":false,"explanation":"A closed circle includes the endpoint and is used for $\\le$ or $\\ge$. Since $\\lt$ and $\\gt$ exclude equality, the correct endpoint is an open circle."},{"text":"A square endpoint.","correct":false,"explanation":"A square endpoint is not the standard notation for inequalities on a number line. Strict inequalities use an open circle because the boundary value is excluded."},{"text":"An arrow endpoint.","correct":false,"explanation":"An arrow shows that a range continues indefinitely, rather than whether a boundary is included. For $\\lt$ or $\\gt$, the endpoint is an open circle."}]},{"id":"2863dfe9-04a0-48e1-8192-95876b567249","cardType":"basic","points":1,"front":"On a number line, what endpoint is used for $\\le$ or $\\ge$?","back":"A closed circle.","mcqOptions":[{"text":"A closed circle.","correct":true,"explanation":"The symbols $\\le$ and $\\ge$ include equality, so the endpoint belongs to the solution set and is represented by a closed circle."},{"text":"An open circle.","correct":false,"explanation":"An open circle excludes the endpoint and is used for strict inequalities. Because $\\le$ and $\\ge$ include equality, the correct endpoint is a closed circle."},{"text":"A square endpoint.","correct":false,"explanation":"A square endpoint is not standard number-line notation for an inclusive inequality. A closed circle shows that the boundary value is included in the solution."},{"text":"An arrow endpoint.","correct":false,"explanation":"An arrow indicates that values continue without bound, not that an endpoint is included. For $\\le$ or $\\ge$, inclusion is shown by a closed circle."}]},{"id":"e65033ce-0105-45d1-b928-2d9fe922d249","cardType":"basic","points":1,"front":"For a solution outside two roots, should the ranges be joined by and or or?","back":"or.","mcqOptions":[{"text":"or.","correct":true,"explanation":"An outside solution consists of two separate ranges, one below the smaller root and one above the larger root. A value can lie in either range, so use “or”."},{"text":"and.","correct":false,"explanation":"Using “and” would require a number to be below the smaller root and above the larger root simultaneously. Outside ranges are alternatives, so they are joined by “or”."},{"text":"both.","correct":false,"explanation":"A single value cannot normally lie in both separated outside ranges at once. It may belong to either outside range, so the correct joining word is “or”."},{"text":"neither.","correct":false,"explanation":"“Neither” would exclude both outside ranges rather than combine them as the solution. Values satisfying an outside inequality lie in one range or the other, so use “or”."}]},{"id":"decc1650-9bf6-4898-9d5e-f7158b1a4ef4","cardType":"basic","points":2,"front":"Solve $x^2+5x+6\\gt 0$.","back":"$x\\lt -3$ or $x\\gt -2$.","mcqOptions":[{"text":"$x\\lt -3$ or $x\\gt -2$.","correct":true,"explanation":"Factoring gives $(x+3)(x+2)\\gt0$. The product is positive outside the roots $-3$ and $-2$, and strict inequality excludes both endpoints."},{"text":"$-3\\lt x\\lt -2$.","correct":false,"explanation":"This selects the interval where $(x+3)$ and $(x+2)$ have opposite signs, making their product negative. The positive solution is $x\\lt-3$ or $x\\gt-2$."},{"text":"$x\\le -3$ or $x\\ge -2$.","correct":false,"explanation":"This correctly identifies the outside regions but mistakenly includes the roots. At $x=-3$ and $x=-2$ the expression equals zero, so strict inequality requires $x\\lt-3$ or $x\\gt-2$."},{"text":"$x\\lt -3$ and $x\\gt -2$.","correct":false,"explanation":"No value can be less than $-3$ and greater than $-2$ simultaneously. The two separate outside ranges must be joined by “or”: $x\\lt-3$ or $x\\gt-2$."}]},{"id":"d3facdac-cddb-49c6-bf48-ce38021c91a8","cardType":"basic","points":2,"front":"Solve $x^2-8x+15\\le 0$.","back":"$3\\le x\\le 5$.","mcqOptions":[{"text":"$3\\le x\\le 5$.","correct":true,"explanation":"Factoring gives $(x-3)(x-5)\\le0$. The product is non-positive between the roots, and equality means the endpoints $3$ and $5$ are included."},{"text":"$x\\le 3$ or $x\\ge 5$.","correct":false,"explanation":"This chooses the regions where both factors have the same sign, making the product positive. The non-positive region lies between the roots: $3\\le x\\le5$."},{"text":"$3\\lt x\\lt 5$.","correct":false,"explanation":"This excludes the roots as though the inequality were strict. Because the expression may equal zero, both roots are included, giving $3\\le x\\le5$."},{"text":"$-5\\le x\\le -3$.","correct":false,"explanation":"This comes from reversing the signs of the roots. The factors are $(x-3)(x-5)$, so the roots are $3$ and $5$, giving $3\\le x\\le5$."}]},{"id":"5cf1f1f6-fd75-4f38-bb44-855b935199dc","cardType":"basic","points":2,"front":"Solve $x^2\\le 64$.","back":"$-8\\le x\\le 8$.","mcqOptions":[{"text":"$-8\\le x\\le 8$.","correct":true,"explanation":"Since $x^2\\le64$, the magnitude of $x$ is at most $8$. Both positive and negative values are allowed, including the endpoints where $x^2=64$."},{"text":"$x\\le -8$ or $x\\ge 8$.","correct":false,"explanation":"This describes values whose squares are at least $64$, reversing the required region. For $x^2\\le64$, the solution is the interval $-8\\le x\\le8$."},{"text":"$-8\\lt x\\lt 8$.","correct":false,"explanation":"This treats the inequality as strict and wrongly excludes $-8$ and $8$. Both endpoints square to $64$, so the correct solution is $-8\\le x\\le8$."},{"text":"$-64\\le x\\le 64$.","correct":false,"explanation":"This uses $64$ directly as the boundary instead of taking its square root. Since $\\sqrt{64}=8$, the correct interval is $-8\\le x\\le8$."}]},{"id":"04ad5f85-1a8c-4bc0-b365-1db510451c7a","cardType":"basic","points":2,"front":"Rearrange $5x+14\\gt x^2$ so one side is zero and the $x^2$ coefficient is positive.","back":"$x^2-5x-14\\lt 0$.","mcqOptions":[{"text":"$x^2-5x-14\\lt 0$.","correct":true,"explanation":"Moving all terms gives $5x+14-x^2\\gt0$. Multiplying by $-1$ changes every sign and reverses the inequality, producing $x^2-5x-14\\lt0$."},{"text":"$x^2-5x-14\\gt 0$.","correct":false,"explanation":"This keeps the inequality direction unchanged after multiplying every term by $-1$. Multiplication by a negative reverses the sign, so the result is $x^2-5x-14\\lt0$."},{"text":"$x^2-5x+14\\lt 0$.","correct":false,"explanation":"The constant sign has not been changed correctly when rearranging. Moving $14$ across, or multiplying by $-1$, makes it $-14$, giving $x^2-5x-14\\lt0$."},{"text":"$x^2+5x-14\\lt 0$.","correct":false,"explanation":"The sign of the linear term is wrong. Moving $5x$ to the side containing $x^2$ makes it $-5x$, so the correct inequality is $x^2-5x-14\\lt0$."}]},{"id":"5cd1cf31-0173-4161-8b11-57687a5ed862","cardType":"basic","points":2,"front":"Solve $5x+14\\gt x^2$.","back":"$-2\\lt x\\lt 7$.","mcqOptions":[{"text":"$-2\\lt x\\lt 7$.","correct":true,"explanation":"Rearranging gives $x^2-5x-14\\lt0=(x-7)(x+2)\\lt0$. An upward-opening quadratic is negative between its roots, so $-2\\lt x\\lt7$."},{"text":"$x\\lt -2$ or $x\\gt 7$.","correct":false,"explanation":"This selects where the upward-opening quadratic is positive, outside its roots. The inequality requires it to be negative, so the solution is between the roots: $-2\\lt x\\lt7$."},{"text":"$-2\\le x\\le 7$.","correct":false,"explanation":"This mistakenly includes the roots, where $5x+14=x^2$ rather than being greater than it. The strict inequality therefore gives $-2\\lt x\\lt7$."},{"text":"$2\\lt x\\lt 7$.","correct":false,"explanation":"This loses the negative sign when solving $x+2=0$. The lower root is $-2$, not $2$, so the correct interval is $-2\\lt x\\lt7$."}]},{"id":"b286b837-c81f-48e7-86e4-5a2ca5fe06b5","cardType":"basic","points":1,"front":"What happens to an inequality sign when you multiply or divide by a negative number?","back":"It reverses.","mcqOptions":[{"text":"It reverses.","correct":true,"explanation":"Multiplying or dividing both sides by a negative number reverses their order on the number line, so the direction of the inequality sign must reverse."},{"text":"It stays the same.","correct":false,"explanation":"Keeping the sign unchanged is valid only when multiplying or dividing by a positive number. With a negative number, the inequality sign reverses."},{"text":"It becomes an equals sign.","correct":false,"explanation":"Multiplication or division cannot turn an inequality into an equation. When the operation uses a negative number, the inequality sign reverses direction."},{"text":"It reverses only when dividing.","correct":false,"explanation":"The sign reversal is not limited to division. Multiplying or dividing both sides by any negative number reverses the direction of the inequality."}]},{"id":"b454fd27-a24f-4626-a68a-a327f45eec39","cardType":"cloze","points":null,"front":"A non-monic quadratic has a coefficient of $x^2$ that is [ ].","back":"A non-monic quadratic has a coefficient of $x^2$ that is not one.","mcqOptions":[{"text":"not one","correct":true,"explanation":"A quadratic is monic when the coefficient of $x^2$ is exactly one, so a non-monic quadratic has an $x^2$ coefficient that is not one."},{"text":"one","correct":false,"explanation":"A coefficient of one makes a quadratic monic, not non-monic. For a quadratic to be non-monic, the coefficient of $x^2$ must be not one."},{"text":"zero","correct":false,"explanation":"If the coefficient of $x^2$ were zero, the expression would no longer be quadratic. A non-monic quadratic instead has an $x^2$ coefficient that is not one."},{"text":"negative","correct":false,"explanation":"A non-monic quadratic may have a negative coefficient, but it need not do so. Its defining feature is that the coefficient of $x^2$ is not one."}]},{"id":"c393586d-8aec-4f2f-bc63-ed056c1056e7","cardType":"basic","points":3,"front":"Solve $6x^2+7x-3\\lt 0$.","back":"$-\\frac{3}{2}\\lt x\\lt \\frac{1}{3}$.","mcqOptions":[{"text":"$-\\frac{3}{2}\\lt x\\lt \\frac{1}{3}$.","correct":true,"explanation":"Factorising gives $(3x-1)(2x+3)\\lt 0$, with roots $-\\frac{3}{2}$ and $\\frac{1}{3}$. The upward-opening quadratic is negative between these roots."},{"text":"$-\\frac{1}{3}\\lt x\\lt \\frac{3}{2}$.","correct":false,"explanation":"This results from assigning the root signs incorrectly. Solving the factors gives $x=-\\frac{3}{2}$ and $x=\\frac{1}{3}$, and the required values lie between them."},{"text":"$\\frac{1}{3}\\lt x\\lt \\frac{3}{2}$.","correct":false,"explanation":"This incorrectly makes both boundary values positive. The factor $2x+3$ gives the negative root $-\\frac{3}{2}$, so the solution is $-\\frac{3}{2}\\lt x\\lt\\frac{1}{3}$."},{"text":"$-\\frac{3}{2}\\lt x\\lt -\\frac{1}{3}$.","correct":false,"explanation":"The second boundary has the wrong sign: $3x-1=0$ gives $x=\\frac{1}{3}$, not $-\\frac{1}{3}$. The quadratic is negative between the two correct roots."}]},{"id":"11dfbe20-e81c-4288-80c4-125d694beb68","cardType":"cloze","points":null,"front":"If a quadratic does not factorise nicely, use the [ ] to find boundary values.","back":"If a quadratic does not factorise nicely, use the quadratic formula to find boundary values.","mcqOptions":[{"text":"quadratic formula","correct":true,"explanation":"The quadratic formula finds the roots of any quadratic equation, including those that do not factorise neatly. These roots provide the boundary values for the inequality."},{"text":"linear formula","correct":false,"explanation":"Linear equations contain no $x^2$ term, so methods for them do not generally find quadratic roots. The quadratic formula gives the required boundary values."},{"text":"gradient formula","correct":false,"explanation":"The gradient formula calculates the slope between points and does not solve quadratic equations. The quadratic formula should be used to find the boundary values."},{"text":"Pythagoras’ theorem","correct":false,"explanation":"Pythagoras’ theorem relates the side lengths of right-angled triangles, rather than finding quadratic roots. The quadratic formula gives the inequality’s boundary values."}]},{"id":"c5746e7b-c599-4e45-b78f-f6c3efa4470c","cardType":"basic","points":3,"front":"What integer values satisfy $x^2-6x+7\\lt 0$?","back":"2, 3 and 4.","mcqOptions":[{"text":"2, 3 and 4.","correct":true,"explanation":"The roots are $3-\\sqrt{2}$ and $3+\\sqrt{2}$, approximately $1.59$ and $4.41$. The upward-opening quadratic is negative between them, giving integers 2, 3 and 4."},{"text":"1, 2, 3 and 4.","correct":false,"explanation":"Including 1 comes from rounding the lower root down rather than respecting the strict interval. Since $1$ lies below $3-\\sqrt{2}$, only 2, 3 and 4 qualify."},{"text":"2, 3, 4 and 5.","correct":false,"explanation":"Including 5 comes from rounding the upper root up rather than checking the strict interval. Since $5$ exceeds $3+\\sqrt{2}$, only 2, 3 and 4 qualify."},{"text":"3 and 4.","correct":false,"explanation":"This omits 2, although substituting it gives $4-12+7=-1$, which is less than zero. The satisfying integer values are 2, 3 and 4."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Quadratic Inequalities","title":"Quadratic Inequalities Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"8699ec85-73e2-4c39-8cc9-e4f80f5dde7e","walled":"$L39"}],"$L3a","$L3b"]