Solve x² + 6x + 9 = 0
Quadratic equation, worked out line by line the way a teacher would write it.
Answer
| Solution | x = −3 |
Step-by-step solution
8 steps-
1 Givenx^{2} + 6 x + 9 = 0
Solve for x.
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2 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = 6 and c = 9, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 6,\quad c = 9 -
3 Factor the trinomial: find two numbers whose product is c = 9 and whose sum is b = 6
The goal is to write x² + bx + c as (x + p)(x + q). Multiplying that out gives x² + (p + q)x + p·q, so p and q must multiply to c = 9 and add up to b = 6.
List the pairs of numbers whose product is 9, with their signs, and pick the pair whose sum is 6.
3 \cdot 3 = 9,\qquad 3 + 3 = 6 -
4 Write the factored form\left(x + 3\right)^{2} = 0
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5 Zero product property: a product is 0 only when one of its factors is 0
0 is the only number with this property: if a·b = 0, then a = 0 or b = 0. That is why the equation was first rearranged to … = 0 and factored: now each factor can be set to 0 on its own, giving a simpler equation for each.
x + 3 = 0 -
6 Subtract 3 from both sides
An equation stays true as long as you do the same thing to both sides, like adding the same weight to both pans of a balance. Adding or subtracting the same amount on both sides moves a term to the other side, where it appears with the opposite sign.
x = -3Move the constant terms to the right.
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7 This factor appears 2 times, so the root has multiplicity 2
A factor that appears more than once, like (x − 2)², gives the same root again. The root is counted that many times; when the count is even, the graph touches the x-axis there instead of crossing it.
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8 Check\begin{aligned}x = -3:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −3 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.