Quadratics

Why quadratics matter in Paper 1 Paper 1

A quadratic is any expression $a{x}^2+bx+c$ with $a\ne 0$. Quadratics run through the whole of Paper 1: they decide where a line meets a curve, how many solutions an equation has, the range of a function, and the shape of a graph. This lesson covers five skills the syllabus lists:

• completing the square, and using the completed square form to find a vertex and sketch a graph;
• the discriminant $b^2-4ac$, and what it tells you about the roots;
• solving quadratic equations and quadratic inequalities;
• solving a linear and a quadratic equation simultaneously, by substitution;
• spotting and solving equations that are quadratic in some function of $x$, such as $x^2$, $\sqrt{x}$ or $\tan x$.

Every paper allows a calculator, but the questions on this topic usually want exact answers (fractions and surds), so the algebra must be secure without one.

Completing the square Paper 1

Completing the square rewrites $a{x}^2+bx+c$ as $a(x+p{)}^2+q$. The idea comes from expanding a perfect square:

$$(x+\frac{b}{2}{)}^2=x^2+bx+\frac{b^2}{4}$$

so $x^2+bx$ is the square $(x+\frac{b}{2}{)}^2$ with $\frac{b^2}{4}$ taken away:

$$x^2+bx=(x+\frac{b}{2}{)}^2-\frac{b^2}{4}$$

Method when $a=1$: halve the coefficient of $x$ to get the number in the bracket, then subtract the square of that number and add the constant. For example $x^2-6x+11=(x-3{)}^2-9+11=(x-3{)}^2+2$.

Method when $a\ne 1$: take $a$ out of the $x^2$ and $x$ terms only, complete the square inside the bracket, then multiply back. For example

$$2x^2+8x-1=2(x^2+4x)-1=2[(x+2{)}^2-4]-1=2(x+2{)}^2-9$$

The $-4$ inside the square bracket is multiplied by 2: this is the step most often lost.

Why the form is useful. A square is never negative, so $(x+p{)}^2⩾0$, with equality only when $x=-p$. So for $y=a(x+p{)}^2+q$:

• if $a>0$, the least value of $y$ is $q$, when $x=-p$: the vertex $(-p,q)$ is a minimum point;
• if $a<0$, the greatest value of $y$ is $q$, when $x=-p$: the vertex is a maximum point;
• the line of symmetry is $x=-p$.

To sketch $y=a{x}^2+bx+c$, show the shape (a U shape if $a>0$, an upside-down U if $a<0$), the vertex from the completed square, the $y$-intercept $(0,c)$, and any $x$-intercepts from solving $a{x}^2+bx+c=0$.

A negative leading coefficient is often written the other way round: $-x^2+6x-4=5-(x-3{)}^2$, which shows the greatest value 5 at once.

The completed square $2(x-3{)}^2+5$ shows the minimum point $(3,5)$ and the line of symmetry $x=3$. Because the least value of $y$ is 5, which is positive, the curve never meets the $x$-axis, so $2x^2-12x+23=0$ has no real roots.

The discriminant Paper 1

Completing the square on $a{x}^2+bx+c=0$ gives the quadratic formula

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

The expression under the square root, $b^2-4ac$, is called the discriminant. It decides how many real roots the equation has, because you can only take the square root of a number that is not negative:

• $b^2-4ac>0$: two distinct (different) real roots, because $\pm$ gives two different values;
• $b^2-4ac=0$: one repeated root (also called equal roots), $x=-\frac{b}{2a}$, because adding and subtracting 0 gives the same value;
• $b^2-4ac<0$: no real roots.

In graph terms, these are the three ways a parabola can sit against the $x$-axis: crossing it twice, touching it at its vertex, or missing it altogether.

"Real roots" (with no word "distinct") means two distinct roots or a repeated root, so the condition is $b^2-4ac⩾0$.

Questions often put an unknown constant $k$ into the coefficients and ask for the values of $k$ that give a certain number of roots. Write down $a$, $b$ and $c$ carefully, including signs and any $k$, form the discriminant, and then solve an equation (for a repeated root) or an inequality (for two roots or none). Often the inequality in $k$ is itself a quadratic inequality, which is solved by the method later in this lesson.

Solving quadratic equations Paper 1

There are three methods, and the syllabus expects all of them.

Factorising is quickest when it works. Rearrange to $a{x}^2+bx+c=0$ first, factorise, and set each factor equal to 0. This works because if a product is 0, one of its factors must be 0. That is also why you must never divide both sides by $x$: dividing by $x$ throws away the solution $x=0$.

Completing the square turns the equation into $a(x+p{)}^2+q=0$, so $(x+p{)}^2=-\frac{q}{a}$ and $x=-p\pm \sqrt{-\frac{q}{a}}$. Remember the $\pm$: a positive number has two square roots.

The quadratic formula always works:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

The formula is not printed in the list of formulae for Paper 1, so learn it.

Exact answers. When a question says "exact" or "in surd form", leave the square root in and simplify it: $\sqrt{40}=\sqrt{4\times 10}=2\sqrt{10}$. A decimal answer then scores no accuracy mark. When it asks for 3 significant figures, use your calculator only at the last step.

Which method? If the discriminant is a perfect square, the quadratic factorises. If a question has just asked you to complete the square, the word "hence" means use that form. Otherwise use the formula.

Quadratic inequalities Paper 1

To solve an inequality such as $x^2-2x-15⩽0$:

1. Rearrange so one side is 0, with the $x^2$ term positive if you can.
2. Solve the equation $x^2-2x-15=0$ to find the critical values: $(x+3)(x-5)=0$, so $x=-3$ and $x=5$.
3. Sketch the graph $y=x^2-2x-15$: a U shape crossing the $x$-axis at $-3$ and $5$.
4. Read off where the graph is in the required position. Here $y⩽0$ means on or below the axis, which is between the roots: $-3⩽x⩽5$.

For a U-shaped graph, "$<0$" is always the region between the roots and "$>0$" is always the two regions outside them. The outside answer has two parts, so it is written as two inequalities joined by "or": $x<-3$ or $x>5$. Never write this as $5<x<-3$, which no number satisfies.

Why sketch? The sketch is the reason for the answer. Students who just "solve" $x^2-2x-15⩽0$ as $x⩽5$ and $x⩽-3$ get it wrong, because the sign of a product depends on both factors together.

Strict or not? If the inequality is $<$ or $>$, the critical values are not included; if it is $⩽$ or $⩾$, they are.

Never multiply or divide by an expression whose sign you do not know. To solve $x^2>4x$, do not divide by $x$; write $x^2-4x>0$, so $x(x-4)>0$, giving $x<0$ or $x>4$.

When a question asks for the values that satisfy two inequalities at once, solve each one, then find the overlap, which a number line shows clearly.

The curve is above the $x$-axis outside the roots, so $3x^2-5x-2>0$ for $x<-\frac{1}{3}$ or $x>2$, and below it between the roots, so $3x^2-5x-2<0$ for $-\frac{1}{3}<x<2$.

Simultaneous equations: one linear, one quadratic Paper 1

When one equation is linear and the other is quadratic (for example a line and a circle, or the syllabus's own example of $2x+3y=7$ with $3x^2=4+4xy$), elimination by adding or subtracting does not work, because the quadratic has terms such as $x^2$ or $xy$ that the line does not. Instead use substitution:

1. Rearrange the linear equation to make $x$ or $y$ the subject. Choose the letter that avoids fractions.
2. Substitute into the quadratic equation, expanding brackets carefully.
3. Solve the resulting quadratic equation in one unknown.
4. Substitute each root back into the linear equation to find the other coordinate. Using the linear one gives exactly one partner for each root; using the quadratic one can produce false pairs.

There are usually two pairs of solutions, and each $x$ must be paired with its own $y$. Write them as pairs, for example $x=2,y=1$ and $x=-6,y=5$.

Geometrically, the solutions are the points where the line meets the curve. The quadratic you get in step 3 can have two, one or no real roots, so the line can cut the curve twice, touch it (a tangent), or miss it. The discriminant of that quadratic tells you which, an idea used heavily in coordinate geometry.

Equations that are quadratic in a function of x Paper 1

Some equations are not quadratics in $x$ but are quadratics in something else. Look for an expression $u$ such that the equation has the form $a{u}^2+bu+c=0$:

• $x^4-5x^2+4=0$ is quadratic in $u=x^2$, because $x^4=(x^2{)}^2$;
• $6x+\sqrt{x}-1=0$ is quadratic in $u=\sqrt{x}$, because $x=(\sqrt{x}{)}^2$;
• ${\tan }^{2}x=1+\tan x$ is quadratic in $u=\tan x$;
• $8x^6+7x^3-1=0$ is quadratic in $u=x^3$, because $x^6=(x^3{)}^2$;
• $\frac{2}{x^2}-\frac{1}{x}=3$ is quadratic in $u=\frac{1}{x}$.

Method: substitute $u$, solve the quadratic in $u$, then go back to $x$ by solving $x^2=u$, $\sqrt{x}=u$, $\tan x=u$ and so on.

The step where marks are lost is going back. Each value of $u$ must be tested:

• $x^2=4$ gives two values, $x=\pm 2$;
• $x^2=-1$ gives no real values;
• $\sqrt{x}=-3$ is impossible, because $\sqrt{x}$ means the positive root;
• $\sin x=2$ is impossible, because $-1⩽\sin x⩽1$, whereas $\tan x$ can take any value.

For trigonometric equations, find every angle in the interval the question gives. $\tan x$ repeats every ${180}^{°}$ (or $\pi$ radians), so each value of $\tan x$ gives two angles between $0^{°}$ and ${360}^{°}$. For $\cos x$ use the symmetry of the graph: $\cos x=c$ has the solutions $\alpha$ and ${360}^{°}-\alpha$ in that interval.