doPhysics.in
All chapters
Algebra for Physics · Lesson 3 of 6

The Quadratic Equation and Its Two Roots

16 min read3 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 3
The quadratic equation and its two roots
Look around
A stone seen twice at one window

A boy on the ground throws a stone straight up at \(20\,\text{m/s}\). His sister watches from a window \(15\,\text{m}\) above him in a Mumbai building. She sees the stone go past her window on the way up. A moment later she sees it again, coming down. The equation of the motion gives two answers for the time at \(15\,\text{m}\), because the stone really is there twice.

you will solve any quadratic of the book and say in advance whether it has two roots, one or none. You will read each root as a moment of the motion.

An equation with a square in it

With upward positive and \(g = 10\,\text{m/s}^2\), the height of the stone after a time \(t\) is \(y = 20t - 5t^2\). Put \(y = 15\) and bring every term to one side: \(5t^2 - 20t + 15 = 0\). Divide by \(5\): \(t^2 - 4t + 3 = 0\). An equation in which the unknown appears squared, and no higher, is a quadratic equation. A value of the unknown that satisfies it is a root.

RememberThe roots of a quadratic
\[ ax^2 + bx + c = 0 \quad\Longrightarrow\quad x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Holds whenfor any numbers \(a\), \(b\), \(c\) with \(a \ne 0\). The roots are real only when \(b^2 - 4ac \ge 0\).
Symbols\(x\) the unknown, \(a\) the number multiplying \(x^2\), \(b\) the number multiplying \(x\), \(c\) the constant term
Units\(b^2\) and \(4ac\) carry the same unit; the roots carry the unit of \(x\)
definition 6.1
Quadratic equation and its roots

A quadratic equation has the form \(ax^2 + bx + c = 0\), with \(a \ne 0\). Its roots are the values of \(x\) that satisfy it. A quadratic has at most two real roots.

Solving by factors

Sometimes the quadratic splits into two factors. \(t^2 - 4t + 3 = (t - 1)(t - 3)\), which you can check by multiplying out. A product is zero only when one of its factors is zero. So either \(t - 1 = 0\) or \(t - 3 = 0\): \(t = 1\,\text{s}\) or \(t = 3\,\text{s}\). The first root is the stone rising past the window, the second is the stone falling past it.

The formula, for every quadratic

Most quadratics of physics do not split into neat factors. The formula above solves every one of them. Its proof is short. It completes a square, so that the unknown appears only once.

Deriving \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) step 1 of 5
1

\[ x^2 + \dfrac{b}{a}x = -\dfrac{c}{a} \]

Why. divide the equation by \(a\), which is not zero, and move the constant term to the right.

Try the formula on the stone. For \(5t^2 - 20t + 15 = 0\), \(a = 5\), \(b = -20\) and \(c = 15\). Then \(t = (20 \pm \sqrt{400 - 300})/10 = (20 \pm 10)/10\), which gives \(3\,\text{s}\) and \(1\,\text{s}\), as the factors did.

The two roots also have a fixed sum and product: \(x_1 + x_2 = -b/a\) and \(x_1x_2 = c/a\). For the stone, \(t_1 + t_2 = 4\,\text{s}\) at every height. So the two passing times always sit symmetrically about \(2\,\text{s}\), the moment of the top. This is the symmetry of rise and fall in the chapter Motion in One Dimension, read off the quadratic.

Checkpoint · numericUnanswered

A ball is thrown straight up at \(25\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). At what time, in seconds, is it \(20\,\text{m}\) above the thrower's hand on the way down?

s± 2 %
CheckpointUnanswered

A stone thrown straight up passes a window at \(1.2\,\text{s}\) and again at \(2.8\,\text{s}\). Take \(g = 10\,\text{m/s}^2\). With what speed was it thrown?

The discriminant counts the roots

Two roots, one, or none

The number under the square root, \(b^2 - 4ac\), is the discriminant. When it is positive, the formula gives two different real roots. When it is zero, the two roots meet in one. When it is negative, there is no real square root and no real root at all. For the stone at a height \(h\), the equation is \(5t^2 - 20t + h = 0\), and the discriminant is \(400 - 20h\).

theorem 6.1
What the discriminant says

Take \(ax^2 + bx + c = 0\) with \(D = b^2 - 4ac\). If \(D > 0\) there are two different real roots. If \(D = 0\) there is one real root, \(x = -b/2a\). If \(D < 0\) there are no real roots.

Move the slider in the figure below and read the discriminant as physics. Below \(20\,\text{m}\) it is positive, and the stone passes the height twice. At exactly \(20\,\text{m}\) it is zero: the stone touches that height once, at the top, at \(2\,\text{s}\). Above \(20\,\text{m}\) it is negative. No real time exists, because the stone never rises that high.

Two roots, two moments at one heightOn the left, a stone thrown straight up at 20 metres per second from the ground, with g = 10; on the right, its height y against time, the parabola y = 20 t minus 5 t squared, from t = 0 to 4 seconds, with a top of 20 metres at 2 seconds. Press play: the stone rises and falls while a dot runs along the parabola at the same height. A slider sets a height h, drawn as a dashed line across both. Where the line cuts the parabola are the roots of 5 t squared minus 20 t plus h = 0, marked with dots and read out. At h = 15 metres the roots are 1 second, going up, and 3 seconds, coming down. As h rises the two roots move together; at 20 metres they merge into one, the top, at 2 seconds; above 20 metres the line misses the parabola and there is no real root, because the stone never gets that high.ball (shaded) 1234510152025Ot (s)y (m) h h = 15 m: t = 1.00 s and 3.00 s
SimA stone thrown up at \(20\,\text{m/s}\) and its height-time graph. Press play to watch the stone and its point on the parabola. Move the slider for the height \(h\). The dashed line cuts the parabola at the two roots of \(5t^2 - 20t + h = 0\). The roots close in as \(h\) rises, merge at the top at \(20\,\text{m}\), and vanish above it.

The largest height with a real root is the maximum height. Setting the discriminant to zero gives it without any other formula: \(400 - 20h = 0\), so \(h = 20\,\text{m}\). In general \(u^2 - 2gh = 0\) gives \(H = u^2/2g\). A zero discriminant often marks a limit like this. A ball just reaches a height, two bodies just touch, or a lens position just exists.

CheckpointUnanswered

A ball is thrown straight up at \(10\,\text{m/s}\). For the time at which it is \(8\,\text{m}\) up, the equation is \(5t^2 - 10t + 8 = 0\). Its discriminant is \(100 - 160 = -60\). What does this tell you?

Which root the physics keeps

Which root to keep

The algebra finds the roots; the physics decides which ones to keep. Three cases come up again and again. Both roots may be real moments, like the two times at the window. One root may be zero, the moment of launch: \(20t - 5t^2 = 0\) is \(t(20 - 5t) = 0\), with roots \(0\) and \(4\,\text{s}\). And one root may be negative, a time before the motion began.

A negative time is not an arithmetic mistake. The equation \(y = ut - \tfrac12 gt^2\) describes a whole parabola, for every time, before the throw as well as after. The ball follows it only from the moment of release. So a negative root is where the same parabola, traced backwards, would have crossed that height. Fig 3.1 shows this for a ball thrown up from a cliff.

The negative root lies on the parabola traced backwardsThe height y of a ball thrown at 20 metres per second, 30 degrees above the horizontal, from the top of a cliff 40 metres high, against time: y = 10 t minus 5 t squared, with y measured upward from the launch point, so the ground below is the dashed line y = minus 40 metres. The flight, from t = 0 to 4 seconds, is drawn solid: the ball rises 5 metres, then falls to the ground at t = 4 seconds. The same parabola continued to earlier times is drawn dashed. It meets the ground line again at t = minus 2 seconds. That second root of t squared minus 2 t minus 8 = 0 belongs to no moment of this throw.1234−1−2−10−20−30−40Ot (s)y (m) 4 s: lands −2 s
Fig 3.1A ball thrown at \(20\,\text{m/s}\), \(30°\) above the horizontal, from a cliff \(40\,\text{m}\) high: its height \(y = 10t - 5t^2\) above the launch point. The solid part is the flight, from \(t = 0\) to the landing at \(4\,\text{s}\). The dashed part is the same parabola before the throw. It meets the ground line at \(t = -2\,\text{s}\), the second root, which the physics rejects.
ExampleThis is the worked example of Motion in Two Dimensions, lesson 5, done with the tool

A ball is thrown at \(20\,\text{m/s}\) at \(30°\) above the horizontal from the top of a cliff \(40\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). Find the time of flight, the distance from the base of the cliff where the ball lands, and the landing speed.

Reveal one step at a time and finish the rest yourself.
CheckpointUnanswered

For a stone thrown up from the edge of a cliff, the time-of-flight equation gives \(t = 5\,\text{s}\) and \(t = -1\,\text{s}\). What is the root \(-1\,\text{s}\)?

Checkpoint · numericUnanswered

In the chapter Motion in a Vertical Circle, a block leaves a smooth hemisphere \(1\,\text{m}\) above the floor, moving at \(2.357\,\text{m/s}\) downward and \(2.108\,\text{m/s}\) sideways. Its fall obeys \(5t^2 + 2.357t - 1 = 0\). Find the time it takes to reach the floor, in seconds.

s± 2 %
Common mistake

A root of the algebra is not always a moment of the physics. Reject a negative time, which lies before the motion began. Reject a root after the motion has ended: the free-fall equation of a stone dropped into a well stops holding at the splash. Never average the two roots to get an answer, except to find the moment of the top. And if \(a = 0\), the equation is not quadratic at all: the formula would divide by zero, so solve the linear equation instead.

In the world
Where this tool works in physics

Quadratics appear wherever a square of time or distance enters. In Motion in One Dimension, a body thrown up passes each height twice, symmetrically about the top. In Motion in Two Dimensions, the time of flight and a throw from a height end in quadratics. So do the equation of the trajectory in its factored form and a launch read from the trajectory. A block leaving a smooth dome lands at the positive root of its fall. The stretch of a bungee cord comes from a quadratic in the extension. The resistance of an endless ladder of resistors is the positive root of a quadratic. The displacement method for a lens finds two positions of the lens as the two roots.

Exam traps
  1. Keeping the negative root, or averaging the two roots.
    Keep only the root that lies inside the motion. The mean of the roots is the moment of the top, not an answer.
  2. Forgetting the \(\pm\) and finding one root only.
    A square root has two signs. Write both roots, then let the physics choose.
  3. Using the formula before the equation is in the form \(ax^2 + bx + c = 0\).
    Bring every term to one side first, and read \(a\), \(b\) and \(c\) with their signs.

Where this tool works in the book

The physics that leans on this lesson. Each card opens the topic where the tool is at work.

Motion in One Dimension
Lesson 7 · Free FallOne sign convention for free fallFree-fall problems need one chosen positive direction.Lesson 7 · Free FallRise and fall are symmetricFor a body thrown vertically upward, the time of rise equals the time of fall, and the speed at any height is the same going up and coming down.
Motion in Two Dimensions
Lesson 4 · Projectile MotionTime of flight height and rangeThe time of flight, maximum height and range follow from the two motions: \(T = 2u\sin\theta / g\), \(H = u^2\sin^2\theta / 2g\), \(R = u^2\sin 2\theta / g\).Lesson 5 · The Trajectory and Its Special CasesThe equation of the trajectoryEliminating time from x(t) and y(t) gives y as a function of x.Lesson 5 · The Trajectory and Its Special CasesProjection from a heightA projectile launched from a height above the landing level is solved with the same two motions; only the vertical displacement is no longer zero.Lesson 6 · The Trajectory as a ToolFactored form of the trajectoryThe trajectory can be written as \(y = x\tan\theta\,(1 - x/R)\).Lesson 6 · The Trajectory as a ToolReading a launch from the trajectoryComparing a given path \(y = Ax - Bx^2\) with the standard trajectory gives \(\tan\theta = A\), the launch speed from \(B\), the range \(A/B\) and the height \(A^2/4B\).
Potential Energy and Conservation of Energy
Lesson 4 · When Friction ActsThe energy account of a bungee jumpIn a bungee jump gravitational energy becomes kinetic energy and then elastic energy.
Motion in a Vertical Circle
Lesson 5 · Leaving a Smooth SphereWhere a particle leaves a sphereThe normal force reaches zero where \(\cos\varphi = 2/3\), about 48.2° from the top, at a height \(2R/3\) above the centre.
Circuits: Kirchhoff's Laws, Networks and RC Transients
Lesson 2 · Symmetry, Bridges and LaddersThe infinite ladderAn infinite ladder contains itself: removing its first section leaves the same ladder, so \(x = a + bx/(b + x)\); with \(a = 1\,\Omega\) and \(b = 2\,\Omega\), \(x = 2\,\Omega\), and with \(a = b = R\), \(x = (1 + \sqrt{5})R/2\).
Spherical Surfaces and Lenses
Lesson 6 · Finding f, and the Faults of LensesThe displacement methodWith object and screen a fixed distance \(D\) apart, a convex lens gives a sharp image at two positions a distance \(d\) apart; then \(f = (D^2 - d^2)/4D\), \(D\) must be at least \(4f\), and the object size is \(\sqrt{I_1I_2}\).