Solve √(x + 41) = x − 1
Radical equation, worked out line by line the way a teacher would write it.
Answer
| Solution | x = 8 |
Step-by-step solution
13 steps-
1 Given\sqrt{x + 41} = x - 1
Solve for x.
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2 Square both sides
Squaring removes a square root, because (√a)² = a. But squaring can also turn a false equation into a true one (−3 ≠ 3, yet 9 = 9), so every answer is checked at the end.
\left(\sqrt{x + 41}\right)^{2} = \left(x - 1\right)^{2}Raising to an even power can create extraneous solutions — every candidate is checked at the end.
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3 Simplify both sidesx + 41 = x^{2} - 2 x + 1
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4 Move every term to the left side so the right side is 0
Factoring and the quadratic formula both work on an equation of the form … = 0. Subtracting the right side from both sides gets there without changing the solutions.
- x^{2} + 3 x + 40 = 0 -
5 Multiply both sides by −1 so the leading coefficient is positive
Multiplying both sides by −1 flips every sign. The equation stays true, and a positive variable or leading term is easier to read and to factor.
x^{2} - 3 x - 40 = 0 -
6 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 = -3 and c = -40, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -3,\quad c = -40 -
7 Factor the trinomial: find two numbers whose product is c = -40 and whose sum is b = -3
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 = -40 and add up to b = -3.
List the pairs of numbers whose product is -40, with their signs, and pick the pair whose sum is -3.
(-8) \cdot 5 = -40,\qquad (-8) + 5 = -3 -
8 Write the factored form\left(x - 8\right) \left(x + 5\right) = 0
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9 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 - 8 = 0\quad \text{or} \quad x + 5 = 0 -
10 Add 8 to 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 = 8Move the constant terms to the right.
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11 Subtract 5 from both sidesx = -5
Move the constant terms to the right.
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12 Check every candidate in the original equation
Each candidate goes back into the original equation. Squaring, or clearing fractions and logarithms, can create extra values that do not work in the original; those extraneous ones are crossed out.
\begin{aligned}x = -5:\quad 6 \neq -6\quad\text{extraneous}\\ x = 8:\quad 7 = 7\quad\checkmark\end{aligned} -
13 Check\begin{aligned}x = 8:\quad 7 = 7\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| 8 | 7 | 7 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.