Mathematics Admissions Test (MAT) Algebra and Equations Syllabus

What is a 'disguised quadratic', and how is it solved?

An equation that becomes quadratic after a substitution. You substitute $u$ for a repeated expression so it takes the form $au^{2}+bu+c=0$, solve for $u$, then back-substitute to find the original variable.

Solve the disguised quadratic $x^{4} - 5x^{2} + 4 = 0$ by substitution.

Let $u = x^{2}$: $u^{2} - 5u + 4 = 0 \Rightarrow (u-1)(u-4)=0$, so $u = 1$ or $u = 4$. Then $x^{2}=1$ or $x^{2}=4$, giving $x = \pm 1, \pm 2$.

What substitution turns $2^{2x} - 5\cdot 2^{x} + 4 = 0$ into a standard quadratic?

Let $u = 2^{x}$ (so $2^{2x} = u^{2}$), giving $u^{2} - 5u + 4 = 0$. Solve for $u$, then solve $2^{x} = u$.

For the equation $x - 5\sqrt{x} + 6 = 0$, what substitution makes it quadratic?

Let $u = \sqrt{x}$ (so $x = u^{2}$), giving $u^{2} - 5u + 6 = 0$. Note $u \geq 0$, so reject any negative $u$ before squaring back.

When solving an equation by squaring both sides (e.g. with surds), what extra step is essential?

Check every solution in the original equation, because squaring can introduce extraneous (false) roots.

What condition on $k$ ensures $x^{2} + kx + 9 = 0$ has two distinct real roots?

Require $\Delta > 0$: $k^{2} - 36 > 0$, i.e. $k > 6$ or $k < -6$.

What condition on the discriminant means a quadratic has a repeated root (the line is tangent / a perfect square)?

$b^{2} - 4ac = 0$.

For what values of $m$ does the line $y = mx + 1$ meet the parabola $y = x^{2}$ exactly once?

Set $x^{2} = mx + 1 \Rightarrow x^{2} - mx - 1 = 0$. One intersection means $\Delta = 0$: $m^{2} + 4 = 0$, which has no real solution, so there is no such real $m$ (the line always cuts the parabola twice).

What does it mean geometrically if the line $y = mx + c$ and parabola $y = ax^{2}+bx+d$ give a quadratic with $\Delta < 0$?

The line and parabola do not intersect (no real points of intersection).

For $ax^{2}+bx+c$ with $a > 0$ to be positive for all real $x$ (always above the axis), what condition is needed?

$a > 0$ and $\Delta = b^{2} - 4ac < 0$ (no real roots, so the curve never touches the $x$-axis).

What is the standard method for solving simultaneous linear equations by elimination?

Scale the equations so one variable has matching coefficients, then add or subtract to eliminate it, solve for the remaining variable, and back-substitute.

How do you solve a linear-and-quadratic pair of simultaneous equations (e.g. a line and a curve)?

Rearrange the linear equation for one variable, substitute it into the quadratic to get a single-variable quadratic, solve it, then substitute back to find the paired values.

When solving one linear and one quadratic equation, why is substitution preferred over elimination?

Because elimination cannot remove the non-linear terms cleanly; substituting the linear relation into the quadratic reduces it to a single quadratic in one variable.

What does the number of real solutions to a line-and-curve simultaneous system tell you geometrically?

It equals the number of intersection points: two solutions = two crossings, one = tangent, zero = no intersection.

When you multiply or divide both sides of an inequality by a negative number, what must you do?

Reverse (flip) the direction of the inequality sign.

What is the general method for solving a quadratic inequality such as $ax^{2}+bx+c > 0$?

Find the roots of $ax^{2}+bx+c=0$, sketch the parabola, then read off the $x$-values where the curve lies above (or below) the $x$-axis as required.

Solve $x^{2} - 5x + 6 < 0$.

Factor: $(x-2)(x-3) < 0$. The parabola opens upward and is negative between its roots, so $2 < x < 3$.

Solve $x^{2} - 5x + 6 > 0$.

Factor: $(x-2)(x-3) > 0$. The expression is positive outside the roots, so $x < 2$ or $x > 3$.

For an upward-opening parabola with roots $\alpha < \beta$, when is $a(x-\alpha)(x-\beta) < 0$ versus $> 0$?

It is negative (below axis) for $\alpha < x < \beta$, and positive (above axis) for $x < \alpha$ or $x > \beta$.

Why can't you 'cross-multiply' an inequality like $\dfrac{1}{x} < 2$ by simply multiplying both sides by $x$?

Because the sign of $x$ is unknown; multiplying by a negative flips the inequality. You must consider the cases $x>0$ and $x<0$ separately (or multiply by $x^{2}>0$).

What is a safe technique for solving an inequality involving a variable in the denominator, e.g. $\dfrac{x+1}{x-2} > 0$?

Determine where numerator and denominator each change sign (critical values $x=-1$ and $x=2$), then test the sign of the whole expression in each interval (excluding values that make the denominator zero).

What key property of squares is used in 'inequalities with algebraic reasoning', e.g. to prove $a^{2}+b^{2} \geq 2ab$?

A square is never negative: $(a-b)^{2} \geq 0$. Expanding gives $a^{2} - 2ab + b^{2} \geq 0$, hence $a^{2} + b^{2} \geq 2ab$.

State the AM-GM inequality for two non-negative numbers $a$ and $b$, and when equality holds.

$\dfrac{a+b}{2} \geq \sqrt{ab}$, with equality if and only if $a = b$.

How do you show $x + \dfrac{1}{x} \geq 2$ for all $x > 0$?

Since $\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^{2} \geq 0$, expanding gives $x - 2 + \dfrac{1}{x} \geq 0$, so $x + \dfrac{1}{x} \geq 2$ (equality at $x=1$). Equivalently, multiply $x+\frac{1}{x}\geq 2$ by $x>0$ to get $x^{2}-2x+1=(x-1)^{2}\geq 0$.

What is the triangle inequality for absolute values?

$|a + b| \leq |a| + |b|$ for all real $a$ and $b$, with equality when $a$ and $b$ have the same sign (or one is zero).

How do you solve an inequality of the form $|x - a| < b$ (with $b>0$)?

It is equivalent to the double inequality $-b < x - a < b$, i.e. $a - b < x < a + b$.

How do you solve $|x - a| > b$ (with $b>0$)?

Split into two cases: $x - a > b$ or $x - a < -b$, giving $x > a + b$ or $x < a - b$.

To prove a strict inequality $A > B$ algebraically, what is a standard strategy?

Consider the difference $A - B$ and show it is always positive, often by writing it as a sum of squares or a product of factors with known signs.

Why is squaring both sides of an inequality only valid under a certain condition?

Squaring preserves the inequality direction only when both sides are non-negative. If either side can be negative, squaring may reverse or invalidate the relation, so signs must be checked first.