Solving an equation and finding where two graphs meet are the same task written two ways. Once you see that, the discriminant tells you how many intersections there are before you solve anything.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Sketch $y = x^2$ and the line $y = 4$ on the same axes. How many times do they meet, and what are the $x$-values? Now do the same for $y = -1$.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
Work through the core explanation before applying it.
Solving $f(x) = k$ means finding the $x$-values where the curve $y = f(x)$ has height $k$, which is exactly where it meets the horizontal line $y = k$. The same idea extends to any two curves: solve them simultaneously and the solutions are their intersection points.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Work through the core explanation before applying it.
If two quadratics are equal for **every** value of $x$, then they are the same expression, so their corresponding coefficients must match: the $x^2$ terms, the $x$ terms and the constants each agree.
So if $ax^2 + bx + c = 3x^2 - 5x + 2$ for all $x$, then $a = 3$, $b = -5$ and $c = 2$. No solving is required, only matching.
The words "for all $x$" matter. If two quadratics are merely equal at some particular values, they are ordinary curves that happen to intersect, and their coefficients need not match at all.
Two quadratics equal for all $x$ have identical coefficients, so match the $x^2$, $x$ and constant terms separately. A missing term means that coefficient is zero. "For all $x$" signals an identity; equality at particular values only means the curves intersect there.
Pause, copy the equating-coefficients rule, a worked match, and the distinction between an identity and an equation, into your book.
We just saw what it means for two expressions to agree everywhere. That raises a question: what does it mean graphically when they agree only at particular values? This card answers it → those values are exactly where the graphs cross.
Solving $f(x) = k$ asks which inputs give output $k$. Graphically that is asking where the curve $y = f(x)$ is at height $k$, which is where it meets the horizontal line $y = k$.
So $x^2 = 4$ has two solutions because $y = x^2$ meets $y = 4$ twice, at $x = -2$ and $x = 2$. And $x^2 = -1$ has no real solutions because the parabola never reaches height $-1$.
This is why a sketch is a legitimate way to answer "how many solutions does this equation have". The count of solutions is the count of intersections.
Solving $f(x) = k$ finds where $y = f(x)$ meets the horizontal line $y = k$, so the number of solutions is the number of intersections. A sketch is therefore a valid way to count solutions.
Pause, copy the equivalence between solving and intersecting, the worked $x^2 = 4$ and $x^2 = -1$ cases, and the note that counting from a sketch is examinable, into your book.
We just saw intersections with a horizontal line. That raises a question: what if the second graph is a sloping line, or another parabola? This card answers it → substitute one into the other, and the discriminant of what results counts the intersections.
Given $y = x^2$ and $y = x + 2$, substitute the linear equation into the quadratic: $x^2 = x + 2$, so $x^2 - x - 2 = 0$, which factorises as $(x-2)(x+1) = 0$.
So $x = 2$ or $x = -1$. Substituting back into the line gives $y = 4$ and $y = 1$, so the intersections are $(2, 4)$ and $(-1, 1)$. Always give both coordinates.
The discriminant of the resulting quadratic counts the intersections before you solve: positive means two, zero means the line is a **tangent** touching at one point, and negative means they never meet.
Substitute the linear equation into the quadratic to get a single quadratic, then solve and substitute back for the $y$-coordinates. Its discriminant counts intersections: two, one (a tangent), or none. For two quadratics, set them equal; the $x^2$ terms may cancel.
Pause, copy the substitution method with the worked $y = x^2$ and $y = x+2$ giving $(2,4)$ and $(-1,1)$, and the three discriminant cases including the tangent, into your book.
Worked examples · 3 in a row, reveal as you go
Follow the reasoning through complete worked solutions.
Find $a$, $b$ and $c$ if $a(x-1)^2 + b(x-1) + c = 2x^2 - 3x + 4$ for all values of $x$.
Find the points of intersection of $y = x^2$ and $y = x + 2$.
Show that $y = 2x - 1$ is a tangent to $y = x^2$, and find the point of contact.
Work through the core explanation before applying it.
Run the quick drill and copy the summary into your book.
At the start you sketched $y = x^2$ against $y = 4$ and against $y = -1$. Explain how the number of intersections in each case matches the number of solutions of the corresponding equation, and what the second case tells you about $x^2 = -1$.
Answer the drill bank and rate your confidence.
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Write full responses, then check them against the model answers.
Q1. Find the points of intersection of $y = x^2 - 2x$ and $y = x + 4$. (3 marks)
Q2. Show that the line $y = 4x - 4$ is a tangent to the parabola $y = x^2$, and state the point of contact. (4 marks)
Q3. Explain why equating coefficients is valid for an identity but not for an ordinary equation, using an example of each. (3 marks)
Practice 1: $a = 5$, $b = 0$, $c = -2$. Practice 2: two, because $y = x^2$ meets $y = 7$ twice, at $x = \pm\sqrt{7}$. Practice 3: $x^2 - 1 = 3$ gives $x = \pm 2$, so $(2,3)$ and $(-2,3)$. Practice 4: $x^2 = 3x$ gives $x(x-3) = 0$, so $(0,0)$ and $(3,9)$.
Q1 (3 marks): $x^2 - 2x = x + 4$, so $x^2 - 3x - 4 = 0$ [1]. $(x-4)(x+1) = 0$, so $x = 4$ or $x = -1$ [1]. $y = 8$ and $y = 3$, giving $(4, 8)$ and $(-1, 3)$ [1].
Q2 (4 marks): $x^2 = 4x - 4$, so $x^2 - 4x + 4 = 0$ [1]. $\Delta = 16 - 16 = 0$ [1]. A zero discriminant means exactly one solution, so the line meets the parabola once and is therefore a tangent [1]. $(x-2)^2 = 0$ gives $x = 2$ and $y = 4$, so contact at $(2, 4)$ [1].
Q3 (3 marks): An identity holds for every value of $x$, so the two expressions are the same polynomial and every corresponding coefficient must agree [1]. For example $a(x+1)^2 = ax^2 + 2ax + a$ for all $x$ lets you match term by term [1]. An ordinary equation such as $x^2 = 4$ holds only at particular values, where the graphs happen to cross, so matching coefficients would be meaningless: the two sides are genuinely different expressions [1].
Take the module quiz if you are ready, then mark the lesson complete.
Match coefficients, count intersections and solve systems at speed. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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