Linear and Quadratic Equations and Inequalities

Lessons

Senior 4 · Mathematics

Linear and Quadratic Equations and Inequalities

5 lessons

Lesson 5 of 5

Systems with quadratic equations, and modelling real problems

25 min

Two facts, two unknowns

A school wants a rectangular vegetable garden with a fence 46 m long all round and an area of $120\,\mathrm{m^2}$ . If the sides are $x$ and $y$ metres, then $2 x + 2 y = 46$ and $x y = 120$ . The second equation is not linear, so this is a system with a quadratic equation in two unknowns. Such systems describe where a line meets a curve, or where two curves meet, and they are the last step in turning a word problem into numbers.

Definition

Solution of a system in $x$ and $y$ : an ordered pair $(x, y)$ that satisfies every equation of the system. Graphically, it is a point where the graphs of the equations meet.

Example: $(3, 4)$ solves $x^2 + y^2 = 25,\ x + y = 7$ , because $9 + 16 = 25$ and $3 + 4 = 7$ .

1. One linear and one quadratic equation: substitute

Make one unknown the subject of the linear equation and substitute it into the quadratic one. The result is a quadratic equation in one unknown, so a line meets a circle or a parabola in 0, 1 or 2 points, according to the sign of its discriminant.

Worked example 1: a line and a circle

Solve $x^{2} + y^{2} = 25$ and $x + y = 7$ .

Solution From the line, $y = 7 - x$ . Substituting:

x^2 + (7 - x)^2 = 25 \iff 2x^2 - 14x + 24 = 0 \iff x^2 - 7x + 12 = 0 \iff (x - 3)(x - 4) = 0 .

$x = 3 \Rightarrow y = 4$ ; $x = 4 \Rightarrow y = 3$ . Solutions: $(3, 4)$ and $(4, 3)$ .

The two solutions of Worked example 1 are the two points where the line $x + y = 7$ cuts the circle $x^{2} + y^{2} = 25$ .

Worked example 2: how many meeting points?

For which values of $k$ does the line $y = x + k$ meet the parabola $y = x^{2}$ in two points, one point, or none?

Solution Substituting: $x^2 = x + k \iff x^2 - x - k = 0$ , with $\Delta = 1 + 4k$ .

$k \gt -\frac{1}{4}$ : $\Delta \gt 0$ , two points.
$k = -\frac{1}{4}$ : $\Delta = 0$ , one point $x = \frac{1}{2}$ , $y = \frac{1}{4}$ ; the line is a tangent.
$k \lt -\frac{1}{4}$ : $\Delta \lt 0$ , no point.

Parallel lines $y = x + k$ and the parabola $y = x^{2}$ : two meeting points ( $k = 2$ ), a tangent ( $k = -\frac{1}{4}$ ) and none ( $k = - 1$ ).

2. Symmetric systems: sum and product

If a system gives the sum $S = x + y$ and the product $P = x y$ , then $x$ and $y$ are the roots of $t^{2} - S t + P = 0$ (Lesson 2). Other symmetric expressions can be turned into $S$ and $P$ : $x^{2} + y^{2} = S^{2} - 2 P$ .

Worked example 3: the garden

Solve the opening problem: $2 x + 2 y = 46$ , $x y = 120$ .

Solution $S = x + y = 23$ and $P = 120$ , so $x$ and $y$ are the roots of $t^{2} - 23 t + 120 = 0$ . $\Delta = 529 - 480 = 49$ , so $t = \frac{23 \pm 7}{2}$ : $t = 15$ or $t = 8$ . The garden is $15\,\mathrm{m}$ by $8\,\mathrm{m}$ .

Check: perimeter $2 (15 + 8) = 46$ ✓, area $15 \times 8 = 120$ ✓.

Worked example 4: both equations quadratic

Solve (a) $x^2 + y^2 = 13,\ xy = 6$ (b) $x^2 + y^2 = 20,\ x^2 - y^2 = 12$ .

Solution

$(x + y)^{2} = x^{2} + y^{2} + 2 x y = 25$ , so $x + y = 5$ or $x + y = - 5$ . With $P = 6$ : $t^{2} - 5 t + 6 = 0$ gives ${2, 3}$ , and $t^{2} + 5 t + 6 = 0$ gives ${- 2, - 3}$ . Four solutions: $(2, 3), (3, 2), (- 2, - 3), (- 3, - 2)$ .
Add the equations: $2 x^{2} = 32$ , so $x = \pm 4$ . Subtract: $2 y^{2} = 8$ , so $y = \pm 2$ . Four solutions: $(4, 2), (4, - 2), (- 4, 2), (- 4, - 2)$ .

Warning

Give solutions as pairs, and match them correctly. In Worked example 1, $(3, 3)$ and $(4, 4)$ are not solutions: each $x$ has its own $y$ , found from the linear equation. Always substitute back into the equation you did not use.

3. Modelling real problems

Name the unknowns, with units.
Translate each fact into an equation or inequality.
Solve.
Reject answers that make no sense in the situation (negative lengths, negative speeds, fractions of people).
Answer the question in words, with units, and check against the original facts.

Worked example 5: a speed problem

A cyclist rides 60 km. Had her average speed been $5\,\mathrm{km\,h^{-1}}$ greater, the ride would have taken one hour less. Find her average speed.

Solution Let the speed be $v\,\mathrm{km\,h^{-1}}$ , $v \gt 0$ . Time = distance ÷ speed, so

\frac{60}{v} - \frac{60}{v + 5} = 1 \iff 60(v + 5) - 60v = v(v + 5) \iff v^2 + 5v - 300 = 0 .

$\Delta = 25 + 1200 = 1225 = 35^2$ , so $v = \frac{-5 \pm 35}{2}$ : $v = 15$ or $v = - 20$ . A speed cannot be negative, so $v = 15\,\mathrm{km\,h^{-1}}$ .

Check: at 15 the ride takes 4 h; at 20 it takes 3 h, one hour less ✓.

Worked example 6: a price range (illustrative figures)

A school canteen finds that when a drink costs $p$ francs it sells $500 - p$ drinks a day (for $0 \lt p \lt 500$ ). For which prices is the daily takings at least 60 000 Frw?

Solution Takings $= p (500 - p)$ . We need

p(500 - p) \ge 60\,000 \iff p^2 - 500p + 60\,000 \le 0 \iff (p - 200)(p - 300) \le 0 ,

so $200 \le p \le 300$ . Any price from 200 to 300 Frw gives at least 60 000 Frw; the most, $250 \times 250 = 62\,500$ Frw, comes at the middle price, 250 Frw.

Tip

In word problems, the quadratic's second root is often a "ghost" (a negative speed, a negative width). Say explicitly why you reject it: examiners award a mark for the reason.

1st Class Maths: Non-linear simultaneous equations (about 15 min). Watch how the linear equation is rearranged and substituted into the quadratic one, and how each x-value is paired with its own y-value.

✦

Practice

Questions 6 to 8 are word problems: define your unknowns first.

Solve $y = x^{2} - 3 x + 1$ and $y = 2 x - 3$ .
Solve $x^{2} + y^{2} = 10$ and $x - y = 2$ .
Solve $x + y = 9$ and $x y = 14$ .
Solve $x^{2} + y^{2} = 34$ and $x y = 15$ .
Find $c$ so that the line $y = 2 x + c$ is a tangent to the parabola $y = x^{2} + 1$ , and find the point of contact.
Two numbers differ by 3 and their product is 108. Find them.
A rectangular plot has a diagonal of 13 m and a perimeter of 34 m. Find its sides.
A bus covers 180 km. If it went $15\,\mathrm{km\,h^{-1}}$ faster, it would take 1 hour less. Find its speed.

Show answersHide answers

$x^{2} - 5 x + 4 = 0$ : $(1, - 1)$ and $(4, 5)$ .
$x = y + 2$ : $2 y^{2} + 4 y - 6 = 0$ , $y = 1$ or $- 3$ : $(3, 1)$ and $(- 1, - 3)$ .
$t^{2} - 9 t + 14 = 0$ : $(2, 7)$ and $(7, 2)$ .
$(x + y)^{2} = 64$ , $x + y = \pm 8$ : $(3, 5), (5, 3), (- 3, - 5), (- 5, - 3)$ .
$x^{2} - 2 x + 1 - c = 0$ , $\Delta = 4 - 4(1 - c) = 4c = 0$ : $c = 0$ ; contact at $x = 1$ , point $(1, 2)$ .
$x(x + 3) = 108 \Rightarrow x^2 + 3x - 108 = 0 \Rightarrow x = 9$ or $- 12$ : the pairs are 9 and 12, or $- 12$ and $- 9$ .
$x + y = 17$ , $x^2 + y^2 = 169 \Rightarrow xy = \frac{289 - 169}{2} = 60$ ; $t^{2} - 17 t + 60 = 0$ : 12 m and 5 m.
$\frac{180}{v} - \frac{180}{v + 15} = 1 \Rightarrow v^2 + 15v - 2700 = 0 \Rightarrow v = 45$ (reject $- 60$ ): $45\,\mathrm{km\,h^{-1}}$ .

Quick check

Two numbers have sum 11 and product 24. Which quadratic equation has them as roots?

Key points

Line and curve: substitute the linear equation into the quadratic one; the discriminant counts the meeting points ( $\Delta = 0$ means a tangent).
Given $x + y = S$ and $x y = P$ , solve $t^{2} - S t + P = 0$ ; use $x^{2} + y^{2} = S^{2} - 2 P$ .
Two quadratics: add, subtract or combine to remove a term, then pair each $x$ with its own $y$ .
In word problems, reject solutions that make no sense and say why.

1Two facts, two unknowns

21. One linear and one quadratic equation: substitute

32. Symmetric systems: sum and product

43. Modelling real problems

Two facts, two unknowns

1. One linear and one quadratic equation: substitute

2. Symmetric systems: sum and product

3. Modelling real problems