Solutions to x² = a (where a > 0) are x = –√a or x = √a — two real, opposite numbers.
What exactly is an x² equation?
A quadratic equation is any equation that can be written as ax² + bx + c = 0, where a ≠ 0, and “quadratic” comes from the Latin word for “square.”
These equations always graph as parabolas because the x² term dominates the shape. A quick example: x² – 5x + 6 = 0, which factors neatly to (x – 2)(x – 3) = 0. Plot y = x² – 5x + 6, and you’ll see the parabola kissing the x-axis right at x = 2 and x = 3.
How do you solve x² − 8x = 24?
The solutions are x = 4 − 2√10 and x = 4 + 2√10.
First, rearrange to standard form: x² − 8x − 24 = 0. Then plug into the quadratic formula with a = 1, b = −8, c = −24. The math works out because (−8)² − 4(1)(−24) = 64 + 96 = 160, and √160 simplifies to 4√10. Plug either solution back in, and the equation balances perfectly.
What do we mean by “solutions of the equation”?
A solution set is the collection of all values that make the equation true when substituted for the unknown.
Take x² = 16 — its solution set is {−4, 4}. Swap in either number, and the equation holds. For something like 3x + 2 = 2x + 5, the only solution is 3. If nothing works, the set is empty, written ∅ or {} as of 2026.
How do you actually solve x² = something?
Move all terms to one side, factor into (x – r₁)(x – r₂) = 0, set each factor to zero, and solve for x.
Start by getting everything on one side: x² + px + q = 0. Then try factoring. If that fails, fall back on the quadratic formula. Always plug your answers back in — it’s the fastest way to catch mistakes.
What’s the trick for solving an equation with two variables?
Isolate one variable, substitute it into the second equation, solve the single-variable equation, then back-substitute to find the other variable.
Say you’ve got y = 2x + 1 and 3x + 2y = 12. Swap the first into the second: 3x + 2(2x + 1) = 12. Solve for x, then drop that value back into y = 2x + 1. You’ll land on x = 2 and y = 5 in no time.
What’s the positive solution of x² − 36 = 5?
The positive solution is x = 9.
Rearrange to x² − 41 = 0, so x² = 41. Square roots give x = ±√41 ≈ ±6.403. Only the positive root counts here, so we round to 6.4 as of 2026.
How do you find the roots of x² = 9?
The roots are x = −3 and x = 3.
Rewrite as x² − 9 = 0, factor to (x + 3)(x − 3) = 0, then use the zero-product rule. Each factor gives one root. Graphically, these are the spots where y = x² − 9 cuts the x-axis.
What’s the role of B in a quadratic?
B is the coefficient of the x term in f(x) = ax² + bx + c, shifting the vertex horizontally and tilting the parabola.
A positive b tugs the vertex left of the y-axis; a negative b shoves it right. Changing b also tweaks the curve’s tilt, flattening or steepening it. Compare f(x) = x² + 6x + 9 (vertex at (−3, 0)) to f(x) = x² − 6x + 9 (vertex at (3, 0)).
How do you complete the square?
Divide by a, move c/a to the right, then add (b/2a)² to both sides and simplify.
Begin with ax² + bx + c = 0. Divide by a to get x² + (b/a)x = −c/a. Add (b/2a)² to both sides; the left side becomes (x + b/2a)², giving you vertex form instantly.
What’s the bare minimum an equation needs?
An equation must have an equals sign connecting two expressions.
Without the “=”, you’ve got an expression, not an equation. The two sides can be numbers (7 = 7) or wild polynomials (x² + 5x − 3 = 2x + 1). The equals sign is what lets you hunt for unknowns.
What does an equation with no solution look like?
An equation with no solution is one whose solution set is empty, such as x = x + 1.
No x can satisfy x = x + 1 because subtracting x from both sides leaves 0 = 1, which never holds. That’s different from an identity like 2x + 2 = 2(x + 1), which has endless solutions.
What are the solutions to the equation x² − 8x = 24?
The solutions are x = 4 − 2√10 and x = 4 + 2√10.
Rearrange to x² − 8x − 24 = 0, then apply the quadratic formula. With a = 1, b = −8, c = −24, you get the two real roots above. Plug them back in and the arithmetic checks out: (−8)² − 4(1)(−24) = 64 + 96 = 160, whose square root is 4√10.
What is the positive solution of x² − 36 = 5?
The positive solution is x = 9.
Rearrange to x² − 41 = 0, so x² = 41. Square roots give x = ±√41 ≈ ±6.403. Only the positive value, 6.403, qualifies as the positive solution, rounding to 6.4 as of 2026.
What are the roots of the quadratic equation x² = 9?
The roots are x = −3 and x = 3.
Rewrite as x² − 9 = 0, factor to (x + 3)(x − 3) = 0, and apply the zero-product property. Each factor set to zero produces one root. Graphically, these points are where the parabola y = x² − 9 crosses the x-axis.
