Mathematics · Expressions, equations and inequalities · Grade 8

The quadratic equation: the parabola

A vegetable plot has an area of 40 m² and a perimeter of 26 m. What are its sides?

The situation

A farmer’s vegetable plot has an area of 40 m² and a fence of 26 m around it. What are its sides? If the width is x, the length is 13 − x, and the area is x(13 − x) = 40 — that is the quadratic equation x² − 13x + 40 = 0. 1200 years ago in Baghdad al-Khwarizmi solved equations like this by “completing” a square. We will complete it too, and then move on to the discriminant and Vieta’s theorem.

Topic

The quadratic equation: incomplete equations, completing the square, the discriminant, Vieta’s theorem, the biquadratic equation, problems. A vegetable plot of area 40 m² and perimeter 26 m: al-Khwarizmi’s completing of the square; the discriminant tells the number of roots

Key ideas

Discriminant
Completing the square
Vieta’s theorem

Formulas

a x^{2} + b x + c = 0, a \neq 0

The standard form of a quadratic equation: $a$ is the leading coefficient, $b$ is the second coefficient, $c$ is the constant term. The roots are the points where the parabola $y = a x^{2} + b x + c$ crosses the $x$ -axis.

x^{2} + b x = c \Rightarrow {(x + \frac{b}{2})}^{2} = c + \frac{b^{2}}{4}

Completing the square (al-Khwarizmi’s method): a square and two equal rectangles, plus the square of the corner.

D = b^{2} - 4 a c, x_{1,2} = \frac{- b \pm \sqrt{D}}{2 a}

The discriminant and the formula for the roots: $D > 0$ — two roots, $D = 0$ — one, $D < 0$ — no real roots.

x_{1} + x_{2} = - \frac{b}{a}, x_{1} x_{2} = \frac{c}{a}

Vieta’s theorem: the sum and the product of the roots can be read from the coefficients; $a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})$ .

Theory

The quadratic equation and its roots

$a x^{2} + b x + c = 0$ ( $a \neq 0$ ) is a quadratic equation. Its root is a number that turns the equation into a true equality. On a graph, $y = a x^{2} + b x + c$ is a parabola; the roots are the points where the parabola crosses the $x$ -axis. The parabola may cross the axis at two points, touch it at one point, or not meet it at all — so there are 2 roots, 1 root or 0 roots. In the vegetable plot problem, $x (13 - x) = 40$ gives $x^{2} - 13 x + 40 = 0$ , with roots 5 and 8.

Incomplete equations

If $b = 0$ or $c = 0$ , the equation is incomplete. For $a x^{2} + b x = 0$ we take $x$ out of the brackets: $x (a x + b) = 0$ , so $x = 0$ or $x = - b / a$ (the parabola passes through the origin). For $a x^{2} + c = 0$ we get $x^{2} = - c / a$ : if the right-hand side is positive, $x = \pm \sqrt{- c / a}$ ; if it is negative, there is no root (a square cannot be negative). For $a x^{2} = 0$ there is a single root $x = 0$ , and the vertex of the parabola is on the axis.

Completing the square: al-Khwarizmi

$x^{2} + 10 x = 39$ : a square with side $x$ and a rectangle of area $10 x$ . We split the rectangle in two ( $5 x$ and $5 x$ ) and place the halves along two sides of the square — a $5 \times 5$ square is missing in the corner. If we add it to both sides: ${(x + 5)}^{2} = 39 + 25 = 64$ , so $x + 5 = 8$ , $x = 3$ . In general, from $x^{2} + b x = c$ we get ${(x + b / 2)}^{2} = c + b^{2} / 4$ . Al-Khwarizmi took only the positive root; today $x + 5 = - 8$ , giving $x = - 13$ , is also called a root.

The discriminant and the formula

If we carry out completing the square in general: from $a x^{2} + b x + c = 0$ we get ${(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$ . The $D = b^{2} - 4 a c$ on the right is the discriminant (“the separator”): if $D > 0$ there are two roots $x_{1,2} = \frac{- b \pm \sqrt{D}}{2 a}$ , if $D = 0$ there is one root $x = - \frac{b}{2 a}$ (the parabola touches the axis), if $D < 0$ there are no real roots (the parabola does not cross the axis). If $b$ is even, the calculation is shorter with $D_{1} = {(b / 2)}^{2} - a c$ .

Vieta’s theorem, biquadratic equations, problems

If the roots are $x_{1}$ , $x_{2}$ , then $a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})$ ; expanding the brackets gives Vieta’s theorem: $x_{1} + x_{2} = - b / a$ , $x_{1} x_{2} = c / a$ . For a monic equation ( $a = 1$ ) the roots can often be found in your head: $x^{2} - 5 x + 6 = 0$ — the sum is 5, the product is 6: these are 2 and 3. In a biquadratic equation $x^{4} + p x^{2} + q = 0$ the substitution $t = x^{2}$ gives a quadratic equation; each positive $t$ gives two roots $\pm \sqrt{t}$ . Problems: area and perimeter (the vegetable plot), motion — if the speed is raised by 10 km/h, a 60 km journey takes 1 hour less: $\frac{60}{v} - \frac{60}{v + 10} = 1$ , hence $v^{2} + 10 v - 600 = 0$ , $v = 20$ .

About the model

In the steps the coefficients are whole numbers; if the roots are fractions or irrational, the formula shows the exact form (a fraction, $\sqrt{D}$ ) and a value rounded with “≈”. The graph is shown from $- 6$ to $6$ — roots outside it are not visible on the axis, but they are in the formula. In the vegetable plot problem a side cannot be negative, so only the positive roots make sense; al-Khwarizmi’s diagram also shows only the positive root.

Did you know?

Muhammad ibn Musa al-Khwarizmi (c. 780–850) worked at the House of Wisdom (Bayt al-Hikma) in Baghdad; in his book “Al-kitab al-mukhtasar fi hisab al-jabr wa-l-muqabala” (c. 820–830) the equation $x^{2} + 10 x = 39$ is solved with a diagram of completing the square. The word “algebra” comes from “al-jabr” in the title of this book.
François Viète (1540–1603), a French lawyer and mathematician, introduced the use of letters for unknown and known quantities in his 1591 work “In artem analyticem isagoge”; the relation between the roots and the coefficients is named after him.

Inside the experiment

The topic is explained step by step with animation, from simple to advanced: from a real-life picture to the formula and the proof. Then you drag the points yourself to check. Inside are missions, a quiz and a guided tour.

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Algebraic fractions — Grade 7
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Systems of equations — Grade 7
Inequalities: ray and interval — Grade 8