Quadratic equations (definition and form)

Standard form. Any quadratic equation in $x$ can be written as

where: $a$ is the coefficient of $x^2$ and must be non-zero (otherwise the $x^2$ term vanishes and the equation collapses to linear), $b$ is the coefficient of $x$ and may be zero, and $c$ is the constant term and may also be zero. The condition $a\neq0$ is not a technicality — it is the entire definition of "quadratic": drop it and you no longer have a degree-2 equation at all.

Getting to standard form. Equations rarely arrive pre-packaged. An equation like $(x-3)(2x+1)=x(x+5)$ has brackets on the left and a product on the right; only after you expand both sides and move every term to one side can you see the true combined degree. Two equations that individually look quadratic can combine so their $x^2$ terms cancel, leaving something linear — so the check $a\neq0$ must always be applied to the final, fully simplified form, never to an intermediate line.

Roots. A number $\alpha$ is a root (or solution) of $ax^2+bx+c=0$ if putting $x=\alpha$ makes the equation true, that is $a\alpha^2+b\alpha+c=0$. Because $p(x)=ax^2+bx+c$ is the same expression you studied earlier as a quadratic polynomial, the roots of $ax^2+bx+c=0$ are exactly the zeroes of that polynomial — solving the equation and finding the zeroes are the same task viewed two ways. A quadratic equation has at most two roots; how many real roots it actually has (two, one repeated, or none) is decided later using the discriminant, a topic this section does not cover.

Where this shows up outside the classroom. The standard-form recognition you are practising here is not just a classroom exercise. The height of a ball thrown upward, written as a function of time, is a quadratic expression; setting it equal to a target height gives a quadratic equation whose roots are the times at which the ball reaches that height. The area of a rectangular plot expressed in terms of one unknown side, once set equal to a target area, likewise becomes a quadratic equation. In every such situation the first task is identical to what this section teaches: simplify fully, check $a\neq0$, and only then move on to solving. Skipping that first check is the single most common reason students misclassify a word problem as quadratic when it is actually linear, or vice versa.

The goal is to take an equation that does not visibly look quadratic, $(x-3)(2x+1)=x(x+5)$, expand both sides fully, collect every term onto one side, and thereby decide with certainty whether it is quadratic and, if so, read off its standard-form coefficients $a$, $b$, $c$ without any algebraic slip along the way.

Step 1. Expand the left side using the distributive rule on the two brackets. $(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$.

Step 2. Expand the right side. $x(x+5) = x^2 + 5x$.

Step 3. Bring every term to one side by subtracting the right side from the left. $2x^2 - 5x - 3 - (x^2+5x) = 0$, which simplifies to $2x^2 - 5x - 3 - x^2 - 5x = 0$, giving $x^2 - 10x - 3 = 0$.

Step 4. Check the leading coefficient against the definition. Here $a=1$, and since $1\neq0$, the equation genuinely is quadratic. Its standard form is $x^2-10x-3=0$, with $a=1,\ b=-10,\ c=-3$. Had the $x^2$ terms cancelled instead — as they would have if the original brackets had been chosen differently — the same procedure would correctly reveal a linear equation instead, which is exactly why the check belongs at the very end, not before expansion.

Determine whether $x=3$ is a root of $2x^2-5x-3=0$, and check whether the equation has a second root at $x=-\tfrac12$.

Step 1. Substitute $x=3$ into the left-hand side of the equation. $2(3)^2 - 5(3) - 3$.

Step 2. Simplify arithmetically. $2(9) - 15 - 3 = 18 - 15 - 3 = 0$.

Step 3. Compare with the right-hand side of the equation, which is $0$. Since the left side also equals $0$, $x=3$ satisfies the equation, so it is a root.

Step 4. Test the candidate second root $x=-\tfrac12$ the same way. $2\left(\tfrac14\right) - 5\left(-\tfrac12\right) - 3 = \tfrac12 + \tfrac52 - 3 = 0$, which also satisfies the equation.

Conclusion: $x=3$ and $x=-\tfrac12$ are both roots of $2x^2-5x-3=0$, confirming that a quadratic equation can be satisfied by two distinct values of $x$, matching the "at most two roots" rule from the core idea.

Competitive stretch

If a problem instead gives you a candidate root and an unknown coefficient — for example, "find $k$ if $x=3$ is a root of $2x^2-5x+k=0$" — the same substitution method solves it directly: $2(9)-5(3)+k=0 \Rightarrow 18-15+k=0 \Rightarrow k=-3$, recovering exactly the equation checked above. Substituting a known root to pin down an unknown constant is a common exam variant of this same core skill.

• Forgetting $a\neq0$. If the $x^2$ coefficient cancels after full simplification, the equation is linear, not quadratic — always check on the fully reduced form.
• Deciding the degree before expanding. $(x-3)(2x+1)=x(x+5)$ looks quadratic on both sides individually, but only expanding and collecting terms reveals the true combined degree.
• Confusing a root with a coefficient. A root is a value of $x$ that makes the equation true; it is not the same object as $a$, $b$, or $c$.
• Stopping after finding one root. A quadratic equation can have up to two roots — verifying only one candidate value does not tell you the equation has no other solution.
• Sign errors while expanding brackets. In $(x-3)(2x+1)$, a slipped sign on the $-3\times2x$ or $-3\times1$ term changes $b$ or $c$ entirely and can silently turn a correct quadratic into a wrong one — expand one term pair at a time and recheck before collecting.

Ask the tutor: Want to practise turning a few tricky-looking equations into standard form, or check your own substitution work for a root — ask the AI tutor to walk through examples with you.