Solve x⁴ − 29x² + 100 = 0
Polynomial equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −5, x = −2, x = 2, x = 5 |
Step-by-step solution
13 steps-
1 Givenx^{4} - 29 x^{2} + 100 = 0
Solve for x.
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2 Every exponent is a multiple of 2: substitute u = x^2
When every exponent is a multiple of 2, the polynomial is really a simpler one in disguise: x⁴ = (x²)², for example. Solve for u = x^2 first, then undo the substitution.
u^{2} - 29 u + 100 = 0 -
3 This is a quadratic in standard form au² + bu + c = 0
Every quadratic equation can be arranged as au² + bu + c = 0. Reading off a = 1, b = -29 and c = 100, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -29,\quad c = 100 -
4 Factor the trinomial: find two numbers whose product is c = 100 and whose sum is b = -29
The goal is to write u² + bu + c as (u + p)(u + q). Multiplying that out gives u² + (p + q)u + p·q, so p and q must multiply to c = 100 and add up to b = -29.
List the pairs of numbers whose product is 100, with their signs, and pick the pair whose sum is -29.
(-25) \cdot (-4) = 100,\qquad (-25) + (-4) = -29 -
5 Write the factored form\left(u - 25\right) \left(u - 4\right) = 0
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6 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.
u - 25 = 0\quad \text{or} \quad u - 4 = 0 -
7 Add 25 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.
u = 25Move the constant terms to the right.
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8 Add 4 to both sidesu = 4
Move the constant terms to the right.
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9 Back-substitute: x^2 = 4x^{2} = 4
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10 Roots of that power equationx = -2\quad \text{or} \quad x = 2
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11 Back-substitute: x^2 = 25x^{2} = 25
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12 Roots of that power equationx = -5\quad \text{or} \quad x = 5
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13 Check\begin{aligned}x = -5:\quad 0 = 0\quad\checkmark\\ x = -2:\quad 0 = 0\quad\checkmark\\ x = 2:\quad 0 = 0\quad\checkmark\\ x = 5:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −5 | 0 | 0 | ✓ |
| −2 | 0 | 0 | ✓ |
| 2 | 0 | 0 | ✓ |
| 5 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.