Find ab and c: Coefficients for the Quadratic Formula, Step by Step

Let the width be $x$ feet. Then the length is $x + 2$ feet. After the renovation both measurements grow by 4 feet, and the new area is 80 square feet.

So $(x + 4)(x + 6) = 80$. Expand: $x^2 + 10x + 24 = 80$. Move the 80 across: $x^2 + 10x - 56 = 0$.

One unknown, one equation — but the unknown is squared, so guessing crawls. The fast exit is the quadratic formula, and its very first step is to find ab and c in $x^2 + 10x - 56 = 0$. The rest of the page teaches you to find ab and c in any quadratic you meet.

Wentworth's classic 1898 algebra states the rule plainly: "Collecting similar terms, every quadratic equation can be made to assume the form $ax^2 + bx + c = 0$." Here $a$, $b$, and $c$ are known numbers, and "the third term $c$ is called the constant term."

To find ab and c, read the equation's three addresses:

• $a$ lives in front of $x^2$ — the square slot;
• $b$ lives in front of $x$ — the linear slot;
• $c$ stands alone — attached to no $x$.

For the flower bed, the slots hold 1, 10, and $-56$: $a = 1$, $b = 10$, $c = -56$. Signs travel with their numbers — keeping signs straight is half of what it means to find ab and c.

Find ab and c in each row, signs included:

A missing term just means a zero coefficient — in $2x^2 - 9 = 0$, $b = 0$. Four rows, four clean find ab and c reads.

The quadratic formula solves every equation of the form $ax^2 + bx + c = 0$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

It comes from completing the square once, in letters: transpose $c$, multiply by $4a$, add $b^2$, extract the root.

Three steps put the formula to work:

1. Find ab and c. Line the equation up in standard form, then read all three coefficients with their signs.
2. Substitute. Drop $a$, $b$, $c$ into the formula — every sign travels along.
3. Simplify. Compute $b^2 - 4ac$ first, then split the $\pm$ into two answers.

Say it as a chant — find ab and c, substitute, simplify — and every quadratic turns routine.

Run the flower bed through. Step 1, find ab and c: $a = 1$, $b = 10$, $c = -56$. Step 2: $x = \frac{-10 \pm \sqrt{100 + 224}}{2}$. Step 3: $\sqrt{324} = 18$, so $x = \frac{-10 \pm 18}{2}$, giving $x = 4$ or $x = -14$.

A width of $-14$ feet is impossible — "the negative root is inapplicable to the problem," as Wentworth says. So the bed is 4 by 6 feet, the enlarged bed is 8 by 10 = 80 square feet, exactly what the picture promised.

Plot $y = x^2 - 2x - 4$, the quadratic behind the disguised equation $x(x - 2) = 4$. The two formula answers, $x = 1 - \sqrt{5}$ and $x = 1 + \sqrt{5}$, land exactly on the x-intercepts. The graph hands you coefficients too: the curve crosses the y-axis at $(0, -4)$, and that height is exactly $c$. And because the parabola opens upward, $a > 0$.

The number under the root, $b^2 - 4ac$, is the discriminant — it "discriminates between the various solution types":

• $b^2 - 4ac > 0$: two real solutions, two x-intercepts — here $4 + 16 = 20$;
• $b^2 - 4ac = 0$: one repeated solution, the curve only kisses the axis — $9x^2 + 12x + 4 = 0$ gives $x = -\frac{2}{3}$;
• $b^2 - 4ac < 0$: no real solutions, no intercepts — $3x^2 + 4x + 2 = 0$ gives $-8$.

Graph or equation, the find ab and c step opens every route. And once you find ab and c, the discriminant already knows how the story ends.

Problem. Solve $3x^2 - 5x + 2 = 0$ — Wentworth's own first formula example from 1898.

Step 1: find ab and c. The equation is already in standard form. Read with signs: $a = 3$, $b = -5$, $c = 2$.

Step 2: substitute. $x = \frac{5 \pm \sqrt{25 - 24}}{6}$. Notice $-b = -(-5) = 5$ — minus a negative flips to plus.

Step 3: simplify. $\sqrt{1} = 1$, so $x = \frac{5 \pm 1}{6}$.

Answer. $x = 1$ or $x = \frac{2}{3}$. Check $x = 1$: $3 - 5 + 2 = 0$. A clean find ab and c pass, exact roots out.

Problem. Solve $3x^2 = 2(1 + 2x)$, an example from Durell's 1911 school algebra.

The equation is not in standard form, so nothing can be read off yet. Expand and move everything left: $3x^2 - 4x - 2 = 0$. Now find ab and c: $a = 3$, $b = -4$, $c = -2$. Tidy, then find ab and c — that order never changes.

Substitute. $x = \frac{4 \pm \sqrt{16 + 24}}{6}$. Watch $-4ac = -4(3)(-2) = +24$ — two negatives multiply to a plus.

Simplify. $x = \frac{4 \pm \sqrt{40}}{6} = \frac{2 \pm \sqrt{10}}{3}$. Leave the radical; as decimals the roots are about $1.721$ and $-0.387$.

Answer. $x = \frac{2 + \sqrt{10}}{3}$ or $x = \frac{2 - \sqrt{10}}{3}$ — found by choosing to find ab and c only after tidying.

Problem. Solve $0 = -3t^2 + 2t + 5$.

Step 1: find ab and c. Keep the given order: $a = -3$, $b = 2$, $c = 5$.

Step 2: substitute. $b^2 - 4ac = 4 - 4(-3)(5) = 64$, so $t = \frac{-2 \pm 8}{2(-3)} = \frac{-2 \pm 8}{-6}$. The denominator $2a = -6$ is negative — signs will flip.

Step 3: simplify. $t = -1$ or $t = \frac{5}{3}$. Check $t = -1$: $-3 - 2 + 5 = 0$. Check $t = \frac{5}{3}$: $-\frac{25}{3} + \frac{10}{3} + 5 = 0$.

A second route: multiply through by $-1$ first, getting $3t^2 - 2t - 5 = 0$. Find ab and c again: $a = 3$, $b = -2$, $c = -5$ — the same two roots fall out. Either order works, just stay consistent.

Every slip below starts the same way — the find ab and c step went wrong.

1. Reading coefficients before standard form. From $x(x - 4) = 21$, do not read $c = 21$. Expand first: $x^2 - 4x - 21 = 0$, so $c = -21$. Find ab and c only after the equation reads "= 0".

2. Dropping the sign of $b$. In $3x^2 - 5x + 2 = 0$, $b = -5$, so $-b = 5$. Writing $-5$ on top hands you wrong roots — the fix is to find ab and c with signs, always.

3. Dividing only part of the top. The $2a$ divides everything above it: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, not $-b \pm \frac{\sqrt{b^2 - 4ac}}{2a}$.

4. Keeping impossible roots. A width of $-14$ feet cannot exist. After you find ab and c and solve, reread the story and discard what the situation forbids.

5. Sign slips inside $b^2 - 4ac$. When $a$ or $c$ is negative, $-4ac$ turns positive: $-4(-3)(5) = +60$. Recompute that product slowly.