A line and a curve can cross twice
Solve the pair of equations and y = x + 2. As with two linear equations, a solution is a pair of values, x and y, that makes both equations true at once, and on a graph it is a point where the two graphs cross.
Two different straight lines cross at most once. But is a curve, a parabola, and the line y = x + 2 crosses it twice. So this pair of equations has two solutions, not one.
The curve is , and the straight line is y = x + 2. They cross in two places.
Substitute the linear equation
At a crossing point, both equations give the same y. So the two expressions for y are equal there: . Replacing y with the expression from the linear equation is substitution, the same move used for two linear equations.
The new equation has only one letter, x, and it has an term, so it is a quadratic equation. Gather every term on one side, so that the other side is 0: .
Solve for x
Factor . Two numbers that multiply to −2 and add to −1 are −2 and 1, so (x − 2)(x + 1) = 0. One of the brackets must be 0, so x = 2 or x = −1.
Find the y that goes with each x
A solution needs a value of y as well as a value of x. Put each x back into the linear equation, y = x + 2, to find its y. When x = 2, y = 2 + 2 = 4. When x = −1, y = −1 + 2 = 1.
Use the linear equation for this step. It is the simpler one, and it gives exactly one y for each x. Then check each pair in the other equation: and , so both pairs make true as well.
The solutions are x = 2, y = 4 and x = −1, y = 1. Keep each x with its own y: x = 2 with y = 1 is not a solution.
The curve and the straight line y = x + 2 cross at (2, 4) and at (−1, 1): these are the two solutions.
When the line is not written as y = …
Solve x + y = 5 and . First rearrange the linear equation so that one letter stands alone: y = 5 − x. Then substitute it into the quadratic: .
Gather on one side: , which factors as (x + 3)(x − 2) = 0. So x = −3 or x = 2. From y = 5 − x, x = −3 gives y = 8, and x = 2 gives y = 3. Check in the quadratic: and .
How many solutions?
A line can cross a parabola twice, touch it once, or miss it altogether, so a pair like this has two solutions, one, or none. The number of real roots of the quadratic equation after substitution is the number of meeting points. The next idea is to count them in advance with the discriminant, without solving.
The usual mistakes
Trying to eliminate x by adding or subtracting the equations, as with two linear equations. Only one of the equations has an term, so there is nothing to cancel it. Put the linear equation into the quadratic instead.
Stopping at x = 2 and x = −1. A solution is a pair of values, so each x needs its y.
The length between two points: Pythagoras' theorem
Part (b) of the next application asks for the length of a straight beam between two points. Going from one point to the other, count how far across and how far up it goes. Those two distances are the two shorter sides of a right-angled triangle, and the beam is its longest side, the hypotenuse.
Pythagoras' theorem says that the squares of the two shorter sides add up to the square of the hypotenuse. For sides 3 and 4, the hypotenuse is .
Worked example: Where a Straight Beam Crosses a Parabolic Arch
Question The arch of a bridge follows the curve y = 6x − x2, where x m is the distance along the ground from the left foot of the arch and y m is the height. A straight steel beam follows the line y = x + 4. (a) Find the coordinates of the two points where the beam meets the arch. (b) Find the length of the beam between these two points, correct to 2 decimal places.
1.At a meeting point both equations hold, so the two expressions for y are equal: x + 4 = 6x − x2.
Where the beam meets the arch both equations hold, so x + 4 = 6x − x2. 2.Add x2 to both sides and subtract 6x from both sides: x2 − 5x + 4 = 0. Its discriminant is b2 − 4ac = 25 − 16 = 9, which is positive, so there are two real roots and the beam meets the arch at two points.
Bring every term to one side: x2 − 5x + 4 = 0. The discriminant is 9, which is positive, so there are two meeting points. 3.Factorize: two numbers with a product of 4 and a sum of −5 are −1 and −4, so (x − 1)(x − 4) = 0, and x = 1 or x = 4.
Factorize: (x − 1)(x − 4) = 0, so x = 1 or x = 4. 4.Substitute each root into y = x + 4: y = 5 when x = 1, and y = 8 when x = 4. (a) The beam meets the arch at (1, 5) and (4, 8). Check in the curve: 6 × 1 − 12 = 5 and 6 × 4 − 42 = 8.
(a) The line gives y = 5 and y = 8, so the beam meets the arch at (1, 5) and (4, 8). 5.(b) From (1, 5) to (4, 8) the beam goes 3 m across and 3 m up. These are the two shorter sides of a right-angled triangle, so by Pythagoras' theorem the length is √32 + 32 = √18 ≈ 4.24 m.
(b) The beam goes 3 m across and 3 m up, so by Pythagoras' theorem its length is √18 ≈ 4.24 m.
Answer: (a) (1, 5) and (4, 8); (b) √18 ≈ 4.24 m
Common mistakes
- Stopping at x = 1 and x = 4. A point has two coordinates, so each root must be substituted back to find its y-coordinate. The linear equation is the easier one to use.
- Adding the two equations as if they were a pair of linear equations. Elimination by adding or subtracting cannot remove x2 here. Substitution works because both equations give y in terms of x.