Solve \( x^2 - 5x + 6 = 0 \).
\( (x-2)(x-3)=0 \) so \( x=2 \) or \( x=3 \).
Algebra & Functions
12 exam-style questions on quadratics, inequalities, factor theorem, graphs and transformations. Tap Show answer for full working.
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Solve the inequality \( 2x - 3 < 5 \) and represent the solution on a number line.
\( 2x < 8 \) ⇒ \( x < 4 \). Open circle at 4, arrow to the left.
Complete the square for \( x^2 + 6x + 5 \) and hence find the minimum value of the expression.
\( x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4 \). Minimum value is \( -4 \) when \( x=-3 \).
Factorise fully \( x^3 - 8 \).
\( x^3 - 8 = (x-2)(x^2 + 2x + 4) \).
Solve \( \dfrac{2x-1}{x+3} = 3 \).
\( 2x-1 = 3(x+3) \) ⇒ \( 2x-1 = 3x + 9 \) ⇒ \( -1-9 = 3x-2x \) ⇒ \( x = -10 \). (Check \( x eq -3 \).)
The quadratic equation \( x^2 + kx + 4 = 0 \) has equal roots. Find the possible values of \( k \).
Discriminant \( = 0 \): \( k^2 - 16 = 0 \) ⇒ \( k = \pm 4 \).
Solve the simultaneous equations \( y = x^2 - 3x + 2 \) and \( y = 2x - 1 \).
\( x^2 - 3x + 2 = 2x - 1 \) ⇒ \( x^2 - 5x + 3 = 0 \) ⇒ \( x = \dfrac{5\pm\sqrt{13}}{2} \). Then \( y = 2x-1 \) for each root.
Given \( f(x) = x^3 - 4x^2 + x + 6 \), show that \( (x-2) \) is a factor and fully factorise \( f(x) \).
\( f(2) = 8 - 16 + 2 + 6 = 0 \), so factor. Synthetic/long division: \( f(x) = (x-2)(x^2 - 2x - 3) = (x-2)(x-3)(x+1) \).
Solve the inequality \( x^2 - 5x + 6 > 0 \).
\( (x-2)(x-3) > 0 \). Critical points 2, 3. Solution: \( x < 2 \) or \( x > 3 \).
The curve \( y = x^2 \) is translated by the vector \( \begin{pmatrix} 2 \\ -3 \end{pmatrix} \). Write down the equation of the new curve and find its minimum point.
New equation: \( y + 3 = (x-2)^2 \) i.e. \( y = (x-2)^2 - 3 \). Minimum at \( (2, -3) \).
Solve \( |2x-1| = |x+3| \).
Cases: \( 2x-1 = x+3 \) ⇒ \( x=4 \); \( 2x-1 = -(x+3) \) ⇒ \( 2x-1 = -x-3 \) ⇒ \( 3x=-2 \) ⇒ \( x=-2/3 \). Both valid.
Sketch the graph of \( y = |x^2 - 4| \) and find the set of values of \( x \) for which \( |x^2 - 4| \leq 5 \).
Graph of \( y=x^2-4 \) reflected above \( x \)-axis where negative. \( |x^2-4|\leq 5 \) ⇒ \( -5 \leq x^2-4 \leq 5 \) ⇒ \( -1 \leq x^2 \leq 9 \). Since \( x^2\geq 0 \): \( 0 \leq x^2 \leq 9 \) ⇒ \( -3 \leq x \leq 3 \).