Topics › KS4 › Algebra · Foundation & Higher

Quadratic equations

quad-01 Present from here

Solve $x^2 + 5x + 6 = 0$

Option A: $x = 2$ or $x = 3$

Option B: $x = -2$ or $x = -3$

Option C: $x = -1$ or $x = -6$

Option D: $x = -1$

Answer and misconceptions

Option A: $x = 2$ or $x = 3$ ✗
Uses the numbers in the brackets as the solutionsFrom $(x + 2)(x + 3) = 0$, gives $2$ and $3$. Solve each bracket: $x + 2 = 0$ gives $x = -2$.

Option B: $x = -2$ or $x = -3$ ✓ Correct

Option C: $x = -1$ or $x = -6$ ✗
Uses a factor pair that multiplies to $6$ but does not add to $5$$(x + 1)(x + 6)$ expands to $x^2 + 7x + 6$. The pair must multiply to $6$ and add to $5$: $2$ and $3$.

Option D: $x = -1$ ✗
Treats $x^2$ as $x$Collects $x^2 + 5x$ as $6x$; $x^2$ and $x$ are not like terms, so factorise instead.
quad-02 Present from here

Solve $x^2 = 3x$

Option A: $x = 3$

Option B: $x = 0$ or $x = -3$

Option C: $x = 3$ or $x = -3$

Option D: $x = 0$ or $x = 3$

Answer and misconceptions

Option A: $x = 3$ ✗
Divides both sides by $x$Dividing by $x$ loses the solution $x = 0$. Rearrange to $x^2 - 3x = 0$ and factorise: $x(x - 3) = 0$.

Option B: $x = 0$ or $x = -3$ ✗
Gets the sign wrong when moving $3x$ acrossWrites $x^2 + 3x = 0$; subtracting $3x$ from both sides gives $x^2 - 3x = 0$.

Option C: $x = 3$ or $x = -3$ ✗
Thinks every quadratic has a positive and a negative solutionCheck: $(-3)^2 = 9$ but $3 \times (-3) = -9$, so $x = -3$ does not work. Factorise $x(x - 3) = 0$ to find the solutions.

Option D: $x = 0$ or $x = 3$ ✓ Correct
quad-03 Present from here

Solve $x^2 - 49 = 0$

Option A: $x = 7$ or $x = -7$

Option B: $x = 7$

Option C: $x = 24.5$

Option D: $x = 49$

Answer and misconceptions

Option A: $x = 7$ or $x = -7$ ✓ Correct

Option B: $x = 7$ ✗
Forgets the negative solution$(-7)^2 = 49$ too. Factorising $(x + 7)(x - 7) = 0$ shows both solutions.

Option C: $x = 24.5$ ✗
Thinks $x^2$ means $2x$Halves $49$; $x^2$ means $x \times x$, so take the square root.

Option D: $x = 49$ ✗
Forgets to square rootStops at $x^2 = 49$; that gives the value of $x^2$, not $x$.
quad-04 Present from here

Which factorisation can be used to solve $x^2 - 16 = 0$?

Option A: $(x - 4)(x - 4)$

Option B: $(x + 8)(x - 8)$

Option C: $(x + 4)(x - 4)$

Option D: $x(x - 16)$

Answer and misconceptions

Option A: $(x - 4)(x - 4)$ ✗
Thinks both brackets take the minus sign$(x - 4)(x - 4) = x^2 - 8x + 16$. For a difference of two squares, one bracket has $+$ and one has $-$, so the $x$ terms cancel.

Option B: $(x + 8)(x - 8)$ ✗
Halves $16$ instead of square-rooting it$(x + 8)(x - 8) = x^2 - 64$. The number in each bracket is $\sqrt{16} = 4$.

Option C: $(x + 4)(x - 4)$ ✓ Correct

Option D: $x(x - 16)$ ✗
Takes out $x$ as a common factor when one term has no $x$$x(x - 16) = x^2 - 16x$. The $16$ has no $x$, so $x$ is not a common factor.
quad-05 Present from here

$2x^2 + 7x + 3 = (2x + 1)(x + 3)$. Solve $2x^2 + 7x + 3 = 0$

Option A: $x = -1$ or $x = -3$

Option B: $x = -\frac{1}{2}$ or $x = -3$

Option C: $x = \frac{1}{2}$ or $x = 3$

Option D: $x = -2$ or $x = -3$

Answer and misconceptions

Option A: $x = -1$ or $x = -3$ ✗
Ignores the coefficient of $x$ in $2x + 1$From $2x + 1 = 0$, gives $x = -1$; that solves $x + 1 = 0$. Here $2x = -1$, so $x = -\frac{1}{2}$.

Option B: $x = -\frac{1}{2}$ or $x = -3$ ✓ Correct

Option C: $x = \frac{1}{2}$ or $x = 3$ ✗
Does not change the sign when moving the number acrossFrom $2x + 1 = 0$, writes $2x = 1$. Subtract $1$ from both sides: $2x = -1$, so $x = -\frac{1}{2}$; likewise $x = -3$.

Option D: $x = -2$ or $x = -3$ ✗
Divides the wrong way roundFrom $2x = -1$, works out $2 \div (-1) = -2$; divide both sides by $2$: $x = -1 \div 2 = -\frac{1}{2}$.