Algebra to TMUA: a short diagnostic

Use $\sqrt{u^2}=|u|$, then decide the sign of $x-3$.

The square root is non-negative, so $\sqrt{(x-3)^2}=|x-3|$. Since $x<3$, we have $x-3<0$ and therefore $|x-3|=-(x-3)$.

Hence $$\frac{\sqrt{(x-3)^2}}{x-3}=\frac{-(x-3)}{x-3}=-1.$$ The denominator is non-zero because $x<3$.

Answer: $-1$

Matched practice

• Modulus and graphs
• Pure Year 2: Functions and Graphs

For $x>4$, simplify $\dfrac{\sqrt{(4-x)^2}}{x-4}$.

Record the zeros of the original denominator before using the sum-of-cubes identity.

The original denominator is zero at $x=3$ and $x=-3$, so both values are excluded.

Factor the numerator and denominator: $$x^3+27=(x+3)(x^2-3x+9),\qquad x^2-9=(x+3)(x-3).$$

For permitted values, cancel the common factor $x+3$. This gives $\dfrac{x^2-3x+9}{x-3}$, with $x\ne -3,3$. The cancellation does not restore $x=-3$ to the original domain.

Answer: $\dfrac{x^2-3x+9}{x-3}$, with $x\ne\pm3$

Matched practice

• Factorisation and useful identities
• Pure Year 1: Algebraic Methods

Simplify $\dfrac{x^3-125}{x^2-25}$, stating all excluded values.

Move everything to one side and use a sign chart. Do not multiply by an expression whose sign you have not fixed.

First exclude $x=-2$. Subtracting $2$ gives $$\frac{x-1}{x+2}-2=\frac{-x-5}{x+2}\geq0.$$

The critical values are $-5$ and $-2$. For $x<-5$, the numerator is positive and the denominator negative. For $-5<x<-2$, both are negative. For $x>-2$, the numerator is negative and the denominator positive.

The fraction is non-negative on $[-5,-2)$. Include $-5$, where the numerator is zero; exclude $-2$, where the expression is undefined.

Answer: $-5\leq x<-2$

Matched practice

• Pure Year 1: Equations and Inequalities
• Logic and proof

Solve $\dfrac{x+2}{x-1}\leq3$.

Solve each condition separately, then take their intersection.

The inequality $x^2>9$ gives $x<-3$ or $x>3$. The second condition is $x<1$.

Every value with $x<-3$ satisfies $x<1$, while no value with $x>3$ does. Therefore the intersection is $x<-3$.

Answer: $x<-3$

Matched practice

• Logic and proof
• Pure Year 1: Equations and Inequalities

Find all real $x$ such that $x^2\leq16$ and $x>-1$.

Split at $x=1/2$, and keep the condition attached to each branch.

If $x\geq1/2$, then $|2x-1|=2x-1$. Thus $2x-1=x+2$, giving $x=3$, which satisfies this branch condition.

If $x<1/2$, then $|2x-1|=1-2x$. Thus $1-2x=x+2$, giving $x=-1/3$, which also satisfies its branch condition.

Both values satisfy the original equation, so the complete solution set is $\{-1/3,3\}$.

Answer: $x=-\dfrac13$ or $x=3$

Matched practice

• Modulus and graph reflections
• Pure Year 2: Functions and Graphs

Solve $|3x+2|=x+4$.

Find which inputs make $f$ negative, then put $2x-1$ between those inputs.

An upward-opening quadratic is negative strictly between its roots, so $f(t)<0$ exactly when $-2<t<4$.

Use $t=2x-1$: $$-2<2x-1<4\quad\Longleftrightarrow\quad -1<2x<5\quad\Longleftrightarrow\quad-\frac12<x<\frac52.$$

The endpoints are excluded because the quadratic is zero there. The condition $f(0)<0$ agrees with the sign of the graph between its roots.

Answer: $-\dfrac12<x<\dfrac52$

Matched practice

• Modulus and graphs
• Pure Year 1: Graphs and Transformations

A quadratic $g$ opens upwards and has roots $1$ and $7$. Find all $x$ such that $g(3x+1)>0$.

Compare the expansions of the sum squared and the difference squared.

Expanding gives $(x+1/x)^2=x^2+2+1/x^2$ and $(x-1/x)^2=x^2-2+1/x^2$.

The second expression is four less than the first. Hence $$(x-1/x)^2=(x+1/x)^2-4=16-4=12.$$ There is no need to solve a quadratic for $x$.

Answer: $12$

Matched practice

• Factorisation and useful identities
• Pure Year 1: Algebraic Expressions

A positive real number $x$ satisfies $x+1/x=5$. Find $(x-1/x)^2$.

Rewrite the equation as two equations for $x^2$, and check the point where one right-hand side becomes zero.

The equation is equivalent to $x^2=4+k$ or $x^2=4-k$. Since $k>0$, the first equation always has two distinct real roots.

When $0<k<4$, the second equation also has two distinct non-zero roots. They differ from the first pair, so there are four roots in total.

When $k=4$, the second equation gives only $x=0$, so there are three distinct roots. When $k>4$, it has no real roots, leaving exactly two.

Answer: $k>4$

Matched practice

• Modulus and graph reflections
• Pure Year 2: Functions and Graphs

For which values of $k>0$ does $|x^2-9|=k$ have exactly three distinct real solutions?