Solve x⁴ − 17x² + 16 = 0
Polynomial equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −4, x = −1, x = 1, x = 4 |
Step-by-step solution
13 steps-
1 Givenx^{4} - 17 x^{2} + 16 = 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} - 17 u + 16 = 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 = -17 and c = 16, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -17,\quad c = 16 -
4 Factor the trinomial: find two numbers whose product is c = 16 and whose sum is b = -17
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 = 16 and add up to b = -17.
List the pairs of numbers whose product is 16, with their signs, and pick the pair whose sum is -17.
(-16) \cdot (-1) = 16,\qquad (-16) + (-1) = -17 -
5 Write the factored form\left(u - 16\right) \left(u - 1\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 - 16 = 0\quad \text{or} \quad u - 1 = 0 -
7 Add 16 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 = 16Move the constant terms to the right.
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8 Add 1 to both sidesu = 1
Move the constant terms to the right.
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9 Back-substitute: x^2 = 1x^{2} = 1
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10 Roots of that power equationx = -1\quad \text{or} \quad x = 1
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11 Back-substitute: x^2 = 16x^{2} = 16
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12 Roots of that power equationx = -4\quad \text{or} \quad x = 4
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13 Check\begin{aligned}x = -4:\quad 0 = 0\quad\checkmark\\ x = -1:\quad 0 = 0\quad\checkmark\\ x = 1:\quad 0 = 0\quad\checkmark\\ x = 4:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| −4 | 0 | 0 | ✓ |
| −1 | 0 | 0 | ✓ |
| 1 | 0 | 0 | ✓ |
| 4 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.