Quadratic equations in one unknown

Quadratic equations are among the oldest problems in mathematics. On Babylonian clay tablets, some of them about $4000$ years old, there are already problems such as "find the sides of a rectangle given its area and the difference of its sides". Around the year $830$ , the mathematician al-Khwarizmi in Baghdad solved such equations by rearranging pieces to "complete a square", and the title of his book (Hisab al-jabr w'al-muqabala) gave us the word "algebra" (MacTutor, University of St Andrews).

The old problem is still close to home. Suppose your family has a rectangular garden whose length is $3$ m more than its width and whose area is $40$ m². How wide is the garden? Let the width be $x$ (m); then $x(x+3)=40$ , that is $x^2+3x-40=0$ . Trying numbers one by one takes time, and the solutions are not always whole numbers. This lesson gives you a formula that solves every equation like this.

Going deeper. The condition $a\neq0$ is essential: if $a=0$ , the equation becomes $bx+c=0$ , a linear equation that you already know how to solve. The coefficients $b$ and $c$ may be $0$ . The equations $x^3-2x=0$ (a cubic) and $\dfrac1x+x=2$ (unknown in a denominator) are not quadratic equations, even though they can be turned into familiar forms.

Solving special quadratic equations

When $c=0$ , we take out the common factor $x$ to get a product equation. When $b=0$ , we rewrite the equation as $x^2=k$ and use square roots.

How many solutions does $x^2=k$ have? Exactly as many as the points where the parabola $y=x^2$ (lesson the function $y=ax^2$ ) meets the horizontal line $y=k$ : two points when $k>0$ , one when $k=0$ , none when $k<0$ .

The number of solutions of

x^2=k

Explore: completing the square

Part c) suggests an idea: if we can write the left-hand side as a square, we can solve the equation. Try it with $x^2-6x+5=0$ . Since $(x-3)^2=x^2-6x+9$ , we add $4$ to both sides:

$x^2-6x+5=0\Leftrightarrow x^2-6x+9=4\Leftrightarrow(x-3)^2=4\Leftrightarrow x-3=\pm2.$

Hence $x=5$ or $x=1$ . This is exactly al-Khwarizmi's "completing the square". Doing the same with any coefficients gives the quadratic formula.

The quadratic formula

Why? Divide both sides by $a$ and complete the square as in the exploration; we get

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}=\frac{\Delta}{4a^2}.$

The right-hand side has denominator $4a^2>0$ , so its sign is the sign of $\Delta$ . This is exactly the form $X^2=k$ above: $\Delta>0$ gives two roots, $\Delta=0$ gives one, and $\Delta<0$ gives none (a square is never negative). The word "discriminant" means the expression that discriminates, or tells apart, the three cases.

Example The three cases of

\Delta

Solve the following equations:

a) $2x^2-5x+2=0$ .

b) $4x^2-4x+1=0$ .

c) $x^2+x+3=0$ .

Solution. a) We have $\Delta=(-5)^2-4\cdot2\cdot2=25-16=9>0$ and $\sqrt\Delta=3$ . The equation has two distinct roots

$x_1=\frac{5+3}{4}=2,\qquad x_2=\frac{5-3}{4}=\frac12.$

b) We have $\Delta=(-4)^2-4\cdot4\cdot1=0$ . The equation has a repeated root $x_1=x_2=\dfrac{4}{2\cdot4}=\dfrac12$ .

c) We have $\Delta=1^2-4\cdot1\cdot3=-11<0$ . The equation has no solutions.

So a) $x=2$ , $x=\dfrac12$ ; b) $x=\dfrac12$ ; c) no solutions.

Now we answer the opening question, going through all four steps of solving a problem.

Example The garden from the opening

A rectangular garden has length $3$ m more than its width and area $40$ m². Find the width and the length of the garden.

Solution.

Understand. We know the difference of the sides ( $3$ m) and the area ( $40$ m²); we need both sides.

Plan. Let the width be $x$ (m), $x>0$ ; the length is $x+3$ (m). The area gives the equation $x(x+3)=40$ , that is $x^2+3x-40=0$ .

Solve. We have $\Delta=3^2-4\cdot1\cdot(-40)=169>0$ and $\sqrt\Delta=13$ . Hence

$x_1=\frac{-3+13}{2}=5,\qquad x_2=\frac{-3-13}{2}=-8.$

Since $x>0$ , $x=5$ . So the garden is $5$ m wide and $8$ m long.

Look back. Check: $5\cdot8=40$ and $8-5=3$ . The root $-8$ is not a length, but it is not meaningless: $x^2+3x-40=(x-5)(x+8)$ , so $-8$ is the "second root" of the same equation. Is there another method? We could look for two numbers with product $40$ that differ by $3$ ; the lesson on Viète's theorem turns this guess into a method.

The reduced discriminant

When $b$ is even, write $b=2b'$ . Then $\Delta=4b'^2-4ac=4(b'^2-ac)$ , and the fractions in the quadratic formula can be simplified by $2$ .

Connections. Quadratic equations tie together many things you have learnt: algebraic identities (to complete the square), square roots (to find the roots), product equations (when $c=0$ ) and the parabola $y=x^2$ (to "see" the number of solutions). Next, Viète's theorem gives the sum and product of the roots without solving, and the lesson solving word problems by forming equations uses the quadratic formula in many real situations.

Key terms

English	Tiếng Việt
quadratic equation in one unknown	phương trình bậc hai một ẩn
coefficient	hệ số
discriminant	biệt thức
quadratic formula	công thức nghiệm
reduced discriminant	biệt thức thu gọn
two distinct roots	hai nghiệm phân biệt
repeated root	nghiệm kép
no (real) solutions	vô nghiệm
completing the square	đưa về bình phương

Summary

A quadratic equation: $ax^2+bx+c=0$ with $a\neq0$ .
$c=0$ : take out the common factor $x$ ; $b=0$ : rewrite as $x^2=k$ .
$\Delta=b^2-4ac$ : $\Delta>0$ two roots $x_{1,2}=\dfrac{-b\pm\sqrt\Delta}{2a}$ ; $\Delta=0$ a repeated root $-\dfrac{b}{2a}$ ; $\Delta<0$ no solutions.
$b=2b'$ : $\Delta'=b'^2-ac$ , $x_{1,2}=\dfrac{-b'\pm\sqrt{\Delta'}}{a}$ .
If $a$ and $c$ have opposite signs, there are always two distinct roots.
Real-life problems: check each root against the condition on the unknown.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Solve the following equations:

a) $5x^2-20=0$ .

b) $-2x^2+6x=0$ .

What you need and how to start

What you need: if $b=0$ , rewrite as $x^2=k$ ; if $c=0$ , take out a common factor.

Steps:

a) Rearrange and divide by $5$ .
b) Write $-2x(x-3)=0$ .

Show solution

a) We have $5x^2=20\Leftrightarrow x^2=4\Leftrightarrow x=2$ or $x=-2$ .

b) We have $-2x(x-3)=0\Leftrightarrow x=0$ or $x=3$ .

So a) $x=\pm2$ ; b) $x=0$ , $x=3$ .

Answer. a) $x=2$ , $x=-2$ ; b) $x=0$ , $x=3$ .

Exercise 2

Solve the following equations using the quadratic formula:

a) $x^2-7x+10=0$ .

b) $2x^2+x-1=0$ .

c) $9x^2-6x+1=0$ .

What you need and how to start

What you need: $\Delta=b^2-4ac$ and the quadratic formula.

Steps:

Identify $a$ , $b$ , $c$ and work out $\Delta$ .
Look at the sign of $\Delta$ , then apply the formula.

Show solution

a) We have $\Delta=49-40=9$ and $\sqrt\Delta=3$ , so $x_1=\dfrac{7+3}{2}=5$ and $x_2=\dfrac{7-3}{2}=2$ .

b) We have $\Delta=1+8=9$ , so $x_1=\dfrac{-1+3}{4}=\dfrac12$ and $x_2=\dfrac{-1-3}{4}=-1$ .

c) We have $\Delta=36-36=0$ , so the equation has a repeated root $x_1=x_2=\dfrac{6}{18}=\dfrac13$ .

So a) $x=5$ , $x=2$ ; b) $x=\dfrac12$ , $x=-1$ ; c) $x=\dfrac13$ .

Answer. a) $5$ ; $2$ . b) $\dfrac12$ ; $-1$ . c) Repeated root $\dfrac13$ .

Exercise 3

Solve the following equations using the reduced discriminant:

a) $x^2-4x-1=0$ .

b) $5x^2+2x+1=0$ .

What you need and how to start

What you need: $b=2b'$ , $\Delta'=b'^2-ac$ .

Steps:

Find $b'$ and work out $\Delta'$ .
Apply $x_{1,2}=\dfrac{-b'\pm\sqrt{\Delta'}}{a}$ .

Show solution

a) We have $b'=-2$ and $\Delta'=4+1=5>0$ , so $x_{1,2}=2\pm\sqrt5$ .

b) We have $b'=1$ and $\Delta'=1-5=-4<0$ , so the equation has no solutions.

So a) $x=2+\sqrt5$ , $x=2-\sqrt5$ ; b) no solutions.

Answer. a) $x=2\pm\sqrt5$ ; b) no solutions.

Exercise 4

Solve the following equations:

a) $(x+1)(x-3)=2x-7$ .

b) $x(x+2)=3(x+4)$ .

What you need and how to start

What you need: expand and rearrange into the form $ax^2+bx+c=0$ .

Steps:

Expand both sides and move everything to the left-hand side.
Solve the resulting quadratic equation.

Show solution

a) We have $x^2-2x-3=2x-7\Leftrightarrow x^2-4x+4=0$ . We have $\Delta'=4-4=0$ , so the equation has a repeated root $x=2$ .

b) We have $x^2+2x=3x+12\Leftrightarrow x^2-x-12=0$ . We have $\Delta=1+48=49$ , so $x_1=\dfrac{1+7}{2}=4$ and $x_2=\dfrac{1-7}{2}=-3$ .

So a) $x=2$ ; b) $x=4$ , $x=-3$ .

Answer. a) $x=2$ ; b) $x=4$ , $x=-3$ .

Exercise 5

Suppose a ball is kicked straight up from the ground and its height after $t$ seconds is $h=-5t^2+20t$ (metres). After how many seconds is the ball at a height of $15$ m?

What you need and how to start

What you need: solve the equation $-5t^2+20t=15$ with $t\geq0$ .

Steps:

Rearrange and divide both sides by $-5$ .
Solve the resulting equation, then explain what the two roots mean.

Show solution

We have $-5t^2+20t=15\Leftrightarrow t^2-4t+3=0$ . We have $\Delta'=4-3=1$ , so $t_1=2+1=3$ and $t_2=2-1=1$ .

So the ball is at a height of $15$ m twice: after $1$ second (on the way up) and after $3$ seconds (on the way down).

Answer. After $1$ second and after $3$ seconds.

Exercise 6

Suppose Mrs Lan puts $100$ million dong into a savings account at an interest rate $r$ per year, and the first year's interest is added to the capital before the second year's interest is worked out (compound interest). After $2$ years she receives $112.36$ million dong. Find the interest rate $r$ .

What you need and how to start

What you need: each year the amount is multiplied by $1+r$ .

Steps:

Form the equation $100(1+r)^2=112.36$ .
Solve it as $X^2=k$ with $X=1+r>0$ .

Show solution

After $2$ years the amount is $100(1+r)^2$ (million dong). We have

$100(1+r)^2=112.36\Leftrightarrow(1+r)^2=1.1236\Leftrightarrow 1+r=1.06\ \text{(since }1+r>0).$

Hence $r=0.06$ .

So the interest rate is $6\%$ per year.

Answer. $r=6\%$ per year.

Exercise 7

Solve the equation $\dfrac{x}{x-2}-\dfrac1x=\dfrac{2}{x(x-2)}$ . (Revise equations with the unknown in a denominator.)

What you need and how to start

What you need: find the conditions for the equation to be defined, clear the denominators and check the conditions.

Steps:

Conditions: $x\neq0$ , $x\neq2$ .
Multiply both sides by $x(x-2)$ .
Solve the resulting quadratic equation and reject any root that breaks the conditions.

Show solution

The equation is defined for $x\neq0$ , $x\neq2$ . Clearing the denominators, we get

$x^2-(x-2)=2\Leftrightarrow x^2-x=0\Leftrightarrow x(x-1)=0\Leftrightarrow x=0\ \text{ or }\ x=1.$

The value $x=0$ breaks the conditions, and $x=1$ satisfies them.

So the equation has the solution $x=1$ .

Answer. $x=1$ .

Exercise 8

(2.0 marks)

a) Solve the equation $2x^2-3x-5=0$ .

b) Find the coordinates of the points where the parabola $(P): y=x^2$ meets the line $(d): y=x+2$ .

c) Solve the equation $x^2-2\sqrt2x-6=0$ .

What you need and how to start

What you need: the quadratic formula; the $x$ -coordinates of the points where $(P)$ meets $(d)$ are the roots of $x^2=x+2$ .

Steps:

a) Work out $\Delta$ .
b) Solve $x^2-x-2=0$ , then work out $y=x^2$ .
c) Use $b'=-\sqrt2$ .

Show solution

a) We have $\Delta=9+40=49$ , so $x_1=\dfrac{3+7}{4}=\dfrac52$ and $x_2=\dfrac{3-7}{4}=-1$ .

b) The $x$ -coordinates of the points of intersection are the roots of $x^2=x+2\Leftrightarrow x^2-x-2=0$ . We have $\Delta=1+8=9$ , so $x=2$ or $x=-1$ . When $x=2$ , $y=4$ ; when $x=-1$ , $y=1$ .

c) We have $b'=-\sqrt2$ , $\Delta'=2+6=8$ and $\sqrt{\Delta'}=2\sqrt2$ . Hence $x_1=\sqrt2+2\sqrt2=3\sqrt2$ and $x_2=\sqrt2-2\sqrt2=-\sqrt2$ .

So a) $x=\dfrac52$ , $x=-1$ ; b) the points of intersection are $(2, 4)$ and $(-1, 1)$ ; c) $x=3\sqrt2$ , $x=-\sqrt2$ .

Answer. a) $\dfrac52$ ; $-1$ . b) $(2, 4)$ , $(-1, 1)$ . c) $3\sqrt2$ ; $-\sqrt2$ .

Exercise 9 Challenge

From a rectangular metal sheet $30$ cm long and $20$ cm wide, four equal squares are cut from the corners and the sides are folded up to make an open box. How long must the side of each square be for the base of the box to have area $336$ cm²? What is the volume of the box then, in cubic centimetres?

What you need and how to start

What you need: the base of the box is a rectangle $(30-2x)\times(20-2x)$ , and the height of the box is $x$ .

Steps:

Let the side of the square be $x$ (cm), $0<x<10$ .
Form and solve the equation $(30-2x)(20-2x)=336$ .
Check against the condition, then find the volume.

Show solution

Let the side of the square be $x$ (cm), $0<x<10$ . From the problem we have the equation

$(30-2x)(20-2x)=336\Leftrightarrow4x^2-100x+264=0\Leftrightarrow x^2-25x+66=0.$

We have $\Delta=625-264=361$ and $\sqrt\Delta=19$ , so $x_1=\dfrac{25+19}{2}=22$ (rejected, since $22>10$ ) and $x_2=\dfrac{25-19}{2}=3$ (satisfies the condition). The volume of the box is then $336\cdot3=1008$ (cm³).

So the side of each square is $3$ cm and the box holds $1008$ cm³.

Answer. $3$ cm; $1008$ cm³.

Mini test

Solutions and answers

Try each exercise on your own first, then check here.

Exercise 1.

a) We have $5x^2=20\Leftrightarrow x^2=4\Leftrightarrow x=2$ or $x=-2$ .

b) We have $-2x(x-3)=0\Leftrightarrow x=0$ or $x=3$ .

So a) $x=\pm2$ ; b) $x=0$ , $x=3$ .

Answer. a) $x=2$ , $x=-2$ ; b) $x=0$ , $x=3$ .

Exercise 2.

a) We have $\Delta=49-40=9$ and $\sqrt\Delta=3$ , so $x_1=\dfrac{7+3}{2}=5$ and $x_2=\dfrac{7-3}{2}=2$ .

b) We have $\Delta=1+8=9$ , so $x_1=\dfrac{-1+3}{4}=\dfrac12$ and $x_2=\dfrac{-1-3}{4}=-1$ .

c) We have $\Delta=36-36=0$ , so the equation has a repeated root $x_1=x_2=\dfrac{6}{18}=\dfrac13$ .

So a) $x=5$ , $x=2$ ; b) $x=\dfrac12$ , $x=-1$ ; c) $x=\dfrac13$ .

Answer. a) $5$ ; $2$ . b) $\dfrac12$ ; $-1$ . c) Repeated root $\dfrac13$ .

Exercise 3.

a) We have $b'=-2$ and $\Delta'=4+1=5>0$ , so $x_{1,2}=2\pm\sqrt5$ .

b) We have $b'=1$ and $\Delta'=1-5=-4<0$ , so the equation has no solutions.

So a) $x=2+\sqrt5$ , $x=2-\sqrt5$ ; b) no solutions.

Answer. a) $x=2\pm\sqrt5$ ; b) no solutions.

Exercise 4.

a) We have $x^2-2x-3=2x-7\Leftrightarrow x^2-4x+4=0$ . We have $\Delta'=4-4=0$ , so the equation has a repeated root $x=2$ .

b) We have $x^2+2x=3x+12\Leftrightarrow x^2-x-12=0$ . We have $\Delta=1+48=49$ , so $x_1=\dfrac{1+7}{2}=4$ and $x_2=\dfrac{1-7}{2}=-3$ .

So a) $x=2$ ; b) $x=4$ , $x=-3$ .

Answer. a) $x=2$ ; b) $x=4$ , $x=-3$ .

Exercise 5.

We have $-5t^2+20t=15\Leftrightarrow t^2-4t+3=0$ . We have $\Delta'=4-3=1$ , so $t_1=2+1=3$ and $t_2=2-1=1$ .

So the ball is at a height of $15$ m twice: after $1$ second (on the way up) and after $3$ seconds (on the way down).

Answer. After $1$ second and after $3$ seconds.

Exercise 6.

After $2$ years the amount is $100(1+r)^2$ (million dong). We have

$100(1+r)^2=112.36\Leftrightarrow(1+r)^2=1.1236\Leftrightarrow 1+r=1.06\ \text{(since }1+r>0).$

Hence $r=0.06$ .

So the interest rate is $6\%$ per year.

Answer. $r=6\%$ per year.

Exercise 7.

The equation is defined for $x\neq0$ , $x\neq2$ . Clearing the denominators, we get

$x^2-(x-2)=2\Leftrightarrow x^2-x=0\Leftrightarrow x(x-1)=0\Leftrightarrow x=0\ \text{ or }\ x=1.$

The value $x=0$ breaks the conditions, and $x=1$ satisfies them.

So the equation has the solution $x=1$ .

Answer. $x=1$ .

Exercise 8.

a) We have $\Delta=9+40=49$ , so $x_1=\dfrac{3+7}{4}=\dfrac52$ and $x_2=\dfrac{3-7}{4}=-1$ .

b) The $x$ -coordinates of the points of intersection are the roots of $x^2=x+2\Leftrightarrow x^2-x-2=0$ . We have $\Delta=1+8=9$ , so $x=2$ or $x=-1$ . When $x=2$ , $y=4$ ; when $x=-1$ , $y=1$ .

c) We have $b'=-\sqrt2$ , $\Delta'=2+6=8$ and $\sqrt{\Delta'}=2\sqrt2$ . Hence $x_1=\sqrt2+2\sqrt2=3\sqrt2$ and $x_2=\sqrt2-2\sqrt2=-\sqrt2$ .

So a) $x=\dfrac52$ , $x=-1$ ; b) the points of intersection are $(2, 4)$ and $(-1, 1)$ ; c) $x=3\sqrt2$ , $x=-\sqrt2$ .

Answer. a) $\dfrac52$ ; $-1$ . b) $(2, 4)$ , $(-1, 1)$ . c) $3\sqrt2$ ; $-\sqrt2$ .

Exercise 9.

Let the side of the square be $x$ (cm), $0<x<10$ . From the problem we have the equation

$(30-2x)(20-2x)=336\Leftrightarrow4x^2-100x+264=0\Leftrightarrow x^2-25x+66=0.$

So the side of each square is $3$ cm and the box holds $1008$ cm³.

Answer. $3$ cm; $1008$ cm³.

Mini test: answers

Part I, question 1: C. $a=-1\neq0$ , $b=0$ , $c=5$ . A and D (with $a=0$ ) are linear equations, and B is a cubic equation.
Part I, question 2: A. $\Delta=9-4\cdot2\cdot(-2)=9+16=25$ . B has the wrong sign of $c$ , C uses $2ac$ instead of $4ac$ , and D writes $b^2=-9$ .
Part I, question 3: B. $\Delta'=9-9=0$ , repeated root $x=-\dfrac{b'}{a}=3$ . A confuses it with $x^2=9$ , C has the wrong sign of $b'$ , and D works out $\Delta'=-9-9<0$ by mistake.
Part I, question 4: A. $b'=-2$ , $\Delta'=4-3\cdot(-1)=7$ . B has the wrong sign of $c$ , C is $\Delta$ , and D uses $b^2$ instead of $b'^2$ .
Part II, question 1: a) True; b) True; c) False; d) False. a) True. b) True: $\Delta'=1+8=9$ . c) False: $x=1\pm3$ , that is $x=4$ , $x=-2$ . d) False: $\Delta'>0$ .
Part III, question 1: 3. $\Delta=1+48=49$ , $x=\dfrac{-1\pm7}{2}$ , that is $x=3$ or $x=-4$ .
Part III, question 2: -0.5. $\Delta=25+24=49$ , $x=\dfrac{5\pm7}{4}$ , that is $x=3$ or $x=-0.5$ .

Quadratic equations in one unknown