Solve √(4x + 29) = 5
Radical equation, worked out line by line the way a teacher would write it.
Answer
| Solution | x = −1 |
Step-by-step solution
8 steps-
1 Given\sqrt{4 x + 29} = 5
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{4 x + 29}\right)^{2} = 5^{2}Raising to an even power can create extraneous solutions — every candidate is checked at the end.
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3 Simplify both sides4 x + 29 = 25
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4 Subtract 29 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.
4 x = -4Move the constant terms to the right.
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5 Divide both sides by 4
Dividing both sides by 4 undoes the multiplication by 4, so the variable is left on its own. Both sides change in the same way, so the equation stays true (dividing by 0 is the one thing that is never allowed).
x = \frac{-4}{4} -
6 Simplifyx = -1
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7 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 = -1:\quad 5 = 5\quad\checkmark\end{aligned} -
8 Check\begin{aligned}x = -1:\quad 5 = 5\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −1 | 5 | 5 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.