Nonlinear Equations in One Variable

Solving a quadratic is a choice among a few reliable routes. Solving an equation with a radical or a fraction is the same, with one extra step that is never optional: checking.

Reading a quadratic was the work of the last two sections; solving one is the work of this one. There are several routes to a quadratic's roots, and a good solver chooses among them rather than forcing a single method onto every problem. Beyond quadratics, two other equation types appear on the test: those with a variable under a radical and those with a variable in a denominator. Both are solved by clearing the obstacle away. That clearing comes with a catch that defines this section: removing a radical by squaring can manufacture a solution that was never really there, and clearing a fraction can let an excluded value slip back in as a false candidate. A final check is part of the method, not a courtesy.

Choose factoring, the square-root method, completing the square, the formula, or graphing

A quadratic equation can be solved four algebraic ways, and they all reach the same roots. Graphing offers a fifth, numerical route: a quick way to see or check the same solutions on Desmos.

• Factoring is fastest when the quadratic factors cleanly. It is the skill from the first section of this domain, finished by the zero-product idea.
• The square-root method fits when there is no middle term, as in $x^2 = k$ or $(x-h)^2 = k$.
• Completing the square always works and is how vertex form is found.
• The quadratic formula works on every quadratic without exception, factorable or not.

Choose the shortest valid method for the form of quadratic in front of you. A quadratic like $x^2 - 5x + 6 = 0$ takes ten seconds to factor into $(x-2)(x-3) = 0$, and the zero-product idea reads the roots straight off the factors: $x = 2$ or $x = 3$. A quadratic like $x^2 - 4x - 3 = 0$ will not factor over the integers, so reach for the formula rather than hunting for a factorization that does not exist.

Solve equations of the form $(x-h)^2 = k$ by square roots

When a quadratic has no middle term, or has been arranged into a perfect square equal to a number, the quickest route is to take the square root of both sides, remembering that a positive number has two square roots, one positive and one negative.

Worked example: solve $(x-3)^2 = 16$. Take the square root of both sides, keeping both signs: $x - 3 = \pm 4$. Then $x = 3 + 4 = 7$ or $x = 3 - 4 = -1$. The $\pm$ is the whole point: dropping the negative root would discard half the answer.

The same move handles a bare squared term. From $2x^2 = 50$, divide to get $x^2 = 25$, then take roots: $x = \pm 5$. A squared variable almost always carries two solutions, and the $\pm$ is how you keep both. When the number is not a perfect square, simplify the radical: the solutions of $x^2 = 12$ are $x = \pm\sqrt{12} = \pm 2\sqrt{3}$, because $\sqrt{12} = \sqrt{4}\cdot\sqrt{3}$.

Solve quadratic equations with the quadratic formula

The quadratic formula is the universal solver: for any equation $ax^2 + bx + c = 0$, it produces the roots directly from the coefficients. It is worth memorizing cold, and worth recognizing that the discriminant from the previous section lives right inside it, under the square root.

Figure 1. The quadratic formula, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with the discriminant $b^2 - 4ac$ sitting under the radical, the same number that counts the roots.

Worked example: solve $2x^2 + 3x - 2 = 0$ with the formula. With $a = 2$, $b = 3$, $c = -2$, substitute into the formula:

$x = \dfrac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \dfrac{-3 \pm \sqrt{9 + 16}}{4} = \dfrac{-3 \pm \sqrt{25}}{4} = \dfrac{-3 \pm 5}{4}$

So $x = \dfrac{2}{4} = \dfrac{1}{2}$ or $x = \dfrac{-8}{4} = -2$. Here the discriminant was 25, a perfect square, so the roots came out rational; this one would have factored too.

The formula earns its keep when a quadratic does not factor nicely. Solving $x^2 - 4x - 3 = 0$, the discriminant is $16 + 12 = 28$, which is not a perfect square, so $x = \dfrac{4 \pm \sqrt{28}}{2} = 2 \pm \sqrt{7}$. These irrational roots are ones no factoring would have found. Two habits prevent most formula errors: write $a$, $b$, $c$ with their signs before substituting, and divide both terms of the numerator by $2a$ when you simplify.

Solve radical equations and check for extraneous solutions

An equation with the variable under a square root is solved by undoing the root: isolate the radical on one side, then square both sides to remove it. But squaring is where the catch lives. It can introduce a solution that satisfies the squared equation yet fails the original. The reason is that squaring forgets signs, so it quietly accepts a reflected version of the equation along with the real one.

Figure 2. Squaring $\sqrt{x} = x - 2$ secretly also solves $\sqrt{x} = 2 - x$. The crossing at $x = 4$ is real; the crossing at $x = 1$ belongs to the reflection, and is extraneous.

Worked example: solve $\sqrt{x} = x - 2$. Square both sides to remove the radical: $x = (x-2)^2 = x^2 - 4x + 4$. Then $x^2 - 5x + 4 = 0$, so $(x-1)(x-4) = 0$ and $x = 1$ or $x = 4$.

Now check both in the original. At $x = 4$: $\sqrt{4} = 2$ and $4 - 2 = 2$, true. At $x = 1$: $\sqrt{1} = 1$ but $1 - 2 = -1$, false. So $x = 4$ is the only solution; $x = 1$ is extraneous, the phantom squaring introduced.

A quick shortcut for spotting the phantom: a square root is never negative, so any candidate that makes the other side negative must fail. But do not skip the substitution; it is the one check that always works.

Solve rational equations and enforce excluded values

An equation with the variable in a denominator is solved by clearing the fractions: multiply every term by the common denominator, which leaves an ordinary equation behind. Often that equation is quadratic. The catch here is the cousin of the radical one: any value that would have made an original denominator zero can never be a solution, no matter what the cleared equation says.

Worked example: solve $x + \dfrac{6}{x} = 5$. Multiply every term by $x$ to clear the fraction: $x^2 + 6 = 5x$, so $x^2 - 5x + 6 = 0$ and $(x-2)(x-3) = 0$. Then $x = 2$ or $x = 3$. Neither makes the denominator $x$ zero, so the check confirms both: $2 + 3 = 5$ and $3 + 2 = 5$. Both are genuine.

Worked example: when the candidate is excluded. Solve $\dfrac{x}{x-1} = \dfrac{1}{x-1} + 2$. Multiply through by $x - 1$: $x = 1 + 2(x-1) = 2x - 1$, so $x = 1$. But $x = 1$ makes the original denominators zero, so it cannot be a solution. The equation has no solution; the only candidate is excluded.

Check candidate solutions created by squaring or denominator clearing

Two operations create extraneous solutions, and they are the two you just used. Squaring both sides can introduce a root from the reflected equation, a genuinely new solution the original relationship never had, because squaring is not reversible without extra sign information. Clearing a denominator works differently: multiplying by the denominator is a valid, reversible step everywhere that denominator is already nonzero, but the resulting equation no longer displays that exclusion, so a value the original forbade can resurface as an apparent candidate.

Either way, for any equation involving a radical or a denominator, substituting your candidates back into the original is the last required step, not an optional flourish. A solution that fails the check is not an error in your algebra; it is a phantom the method was always going to produce, and discarding it is part of solving correctly.

Use the zero-product property to preserve every solution

An equation with every term sharing a factor of $x$ invites a tempting shortcut: dividing both sides by $x$. That quietly destroys a solution. Dividing by $x$ assumes $x$ is not zero, but $x = 0$ is often exactly one of the roots the equation has.

Worked example: keep the root that division would erase. What is the product of the distinct solutions to $x^3 - 7x^2 + 10x = 0$? Factor out the shared $x$ instead of dividing it away: $x(x^2 - 7x + 10) = 0$, so $x(x-5)(x-2) = 0$. The three solutions are $x = 0$, $5$, and $2$. Their product is $0$. Dividing the original equation by $x$ at the start would have silently discarded the root $x = 0$ and produced only $5$ and $2$, missing the zero entirely.

Count solutions to a system from graph intersections

Not every system is two straight lines. A line can meet a parabola, or a circle, and the governing idea does not change: the solutions are the places where both equations hold at once. You find them by substituting one equation into the other, which collapses the pair into a single equation. When that equation is quadratic, the number of solutions is the number of times the two graphs cross.

Figure 3. A line can cut a parabola twice, touch it once (tangent), or miss it entirely. The count is the number of shared solutions.

The number of crossings is read from the resulting quadratic the same way you read any quadratic's roots, through its discriminant. Two real roots means two intersections; one repeated root means the line is tangent, touching at a single point; no real root means the graphs never meet.

Worked example: a tangent condition. For what value of $c$ is the line $y = 6x + c$ tangent to $y = x^2$? Tangent means exactly one shared point. Set the expressions equal and gather to one side: $x^2 = 6x + c$, so $x^2 - 6x - c = 0$. One repeated root requires the discriminant to be zero: $(-6)^2 - 4(1)(-c) = 0$, so $36 + 4c = 0$ and $c = -9$. At $c = -9$ the line just grazes the parabola at a single point.

Figure 4. A tangent line meets the curve exactly once, the nonlinear face of "one solution."

The discriminant itself gets its full treatment in the preceding quadratic-equations chapter; here it is simply the instrument that counts how many times two graphs can agree. The lesson that carries is the framing: a nonlinear system is still a question about where the restrictions meet, and substituting reduces that question to one you can already read.

A nonlinear system inside a word problem

A ball thrown from ground level follows the path $h = -x^2 + 8x$, where $h$ is height and $x$ is horizontal distance, both in feet. A drone holds a constant height of 12 feet along the same path. At what horizontal distances is the ball level with the drone?

The word problem is a system in disguise: the parabola $h = -x^2 + 8x$ and the horizontal line $h = 12$. Set them equal, exactly as before: $-x^2 + 8x = 12$, so $x^2 - 8x + 12 = 0$ and $(x-2)(x-6) = 0$. The ball reaches 12 feet twice, at $x = 2$ and $x = 6$, once on the way up and once on the way down. Both solutions are valid here since neither horizontal distance is negative. A context can rule out a root the algebra alone would accept just as easily as it accepted these two.

Chapter summary

Solving a quadratic is choosing the lightest route: factoring when it is clean, the square root when there is no middle term, and the formula when nothing else fits. The formula, which carries the discriminant inside it, never fails. Radical and rational equations add one obstacle and one discipline. Clear the radical by squaring or the fraction by multiplying, then check every candidate in the original, because squaring can introduce extraneous roots and denominator clearing can leave excluded values among the candidates. A solution that survives the check is real; one that fails it was never there. Keeping that final step is the difference between solving and merely rearranging.