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How to Write the Quadratic Equation Whose Roots Are Given

·9 min read·Solvify Team

To write the quadratic equation whose roots are given, you reverse the usual solving process: instead of extracting roots from an equation, you construct the equation from its roots. The method rests on a single idea — if r₁ and r₂ are roots of a quadratic, then (x − r₁)(x − r₂) = 0. This guide covers every case you will encounter, from whole-number roots to fractions, irrational numbers, and complex conjugates, each illustrated with fully worked examples and self-check steps.

What Does It Mean to Write a Quadratic Equation from Its Roots?

A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. Its roots (also called zeros or solutions) are the x-values that satisfy the equation. When a problem asks you to write the quadratic equation whose roots are, say, 3 and 5, it is asking you to work backwards — find an equation that produces exactly those two roots when solved. This is a core algebra skill tested from Algebra 2 through pre-calculus, and it connects directly to factoring, graphing parabolas, and building higher-degree polynomials. The key insight is that roots and factors are two sides of the same coin: if x = r is a root, then (x − r) is a factor of the quadratic.

Every quadratic with roots r₁ and r₂ can be written as a(x − r₁)(x − r₂) = 0, where a is any nonzero constant — usually taken as 1 unless the problem states otherwise.

The Factor Form Method — Step by Step

The most direct approach is to use the factored form. Because a root is a value that makes a factor equal to zero, the two factors must be (x − r₁) and (x − r₂). Multiplying these factors and expanding gives the standard-form equation. This three-step process works for any pair of real roots, regardless of sign or size. Work through the sign substitution carefully — it is the step where most errors occur.

1. Step 1 — Write the factored form

Start with (x − r₁)(x − r₂) = 0. Substitute the given root values for r₁ and r₂, paying close attention to signs. For roots 3 and 5: (x − 3)(x − 5) = 0.

2. Step 2 — Expand using FOIL

Multiply the two binomials. (x − 3)(x − 5) = x·x + x·(−5) + (−3)·x + (−3)·(−5) = x² − 5x − 3x + 15 = x² − 8x + 15.

3. Step 3 — Write in standard form and verify

Set the expanded expression equal to zero: x² − 8x + 15 = 0. This is the quadratic equation with roots 3 and 5. Verify by substituting: x = 3 → 9 − 24 + 15 = 0 ✓. x = 5 → 25 − 40 + 15 = 0 ✓.

Vieta's Formulas — The Sum and Product Shortcut

Vieta's formulas offer a faster route that skips the expansion step entirely. For a monic quadratic x² + bx + c = 0 (leading coefficient 1), the sum of the roots equals −b and the product of the roots equals c. Rearranged, this gives the template x² − (sum of roots)x + (product of roots) = 0. Vieta's formulas are especially useful when you need to write the quadratic equation whose roots are given as algebraic expressions rather than specific numbers, or when you want to check a factoring result quickly.

1. Step 1 — Find the sum of the roots

Add the two roots. Example: roots are −2 and 7. Sum = −2 + 7 = 5.

2. Step 2 — Find the product of the roots

Multiply the two roots. Product = (−2) × 7 = −14.

3. Step 3 — Substitute into the Vieta template

x² − (sum)x + (product) = 0 becomes x² − 5x + (−14) = 0, which simplifies to x² − 5x − 14 = 0.

4. Step 4 — Verify by factoring

x² − 5x − 14 factors as (x − 7)(x + 2) = 0, giving roots x = 7 and x = −2 ✓.

For any monic quadratic x² + bx + c = 0: sum of roots = −b and product of roots = c.

Worked Examples with Integer Roots

Integer roots are the most common type on quizzes and standardized tests. The four examples below cover positive roots, mixed signs, both-negative roots, and a zero root — each scenario produces a predictable sign pattern in the resulting equation. Recognizing these patterns helps you write and check equations faster.

1. Example 1 — Both roots positive: roots 4 and 6

Sum = 4 + 6 = 10. Product = 4 × 6 = 24. Equation: x² − 10x + 24 = 0. Both the middle term and constant are positive when both roots are positive. Check: (x − 4)(x − 6) = x² − 10x + 24 ✓.

2. Example 2 — Mixed signs: roots −3 and 8

Sum = −3 + 8 = 5. Product = (−3) × 8 = −24. Equation: x² − 5x − 24 = 0. The constant is negative when the roots have opposite signs. Check: (x + 3)(x − 8) = x² − 8x + 3x − 24 = x² − 5x − 24 ✓.

3. Example 3 — Both roots negative: roots −5 and −2

Sum = −5 + (−2) = −7. Product = (−5)(−2) = 10. Equation: x² − (−7)x + 10 = x² + 7x + 10 = 0. Both terms are positive because two negatives multiply to a positive. Check: (x + 5)(x + 2) = x² + 7x + 10 ✓.

4. Example 4 — One root is zero: roots 0 and 9

Sum = 0 + 9 = 9. Product = 0 × 9 = 0. Equation: x² − 9x + 0 = 0, which simplifies to x² − 9x = 0. Check: x(x − 9) = 0 gives x = 0 or x = 9 ✓.

When both roots are negative, both the middle-term coefficient and the constant are positive — the opposite pattern from both-positive roots.

Worked Examples with Fraction and Irrational Roots

Fraction roots and irrational roots appear on standardized tests and in precalculus. With fraction roots, it is often cleaner to clear denominators by multiplying through by the LCM after applying Vieta's formulas. Irrational roots almost always come in conjugate pairs of the form a + √b and a − √b, which is convenient: the surds cancel in the sum, and the product becomes a difference of squares with no radicals left.

1. Example 1 — Fraction roots: 1/2 and 3/4

Sum = 1/2 + 3/4 = 2/4 + 3/4 = 5/4. Product = (1/2)(3/4) = 3/8. Base equation: x² − (5/4)x + 3/8 = 0. Multiply every term by 8 to clear fractions: 8x² − 10x + 3 = 0. Verify: discriminant = 100 − 96 = 4, roots = (10 ± 2)/16 = 3/4 or 1/2 ✓.

2. Example 2 — Pure surd roots: √5 and −√5

Sum = √5 + (−√5) = 0. Product = (√5)(−√5) = −5. Equation: x² − 0·x + (−5) = 0 → x² − 5 = 0. Check: x² = 5, x = ±√5 ✓.

3. Example 3 — Conjugate surd roots: 2 + √3 and 2 − √3

Sum = (2 + √3) + (2 − √3) = 4. Product = (2 + √3)(2 − √3) = 4 − 3 = 1. Equation: x² − 4x + 1 = 0. Check: quadratic formula gives x = (4 ± √(16 − 4))/2 = (4 ± √12)/2 = 2 ± √3 ✓.

Conjugate surd roots (a ± √b) always produce a quadratic with integer coefficients — their sum and product are both rational numbers.

Writing Quadratics with Complex Roots

Complex roots always appear as conjugate pairs: if one root is a + bi, the other is a − bi (where i = √(−1)). This is guaranteed by the complex conjugate root theorem for polynomials with real coefficients. The algebra is identical to the surd case — use Vieta's formulas and the imaginary parts cancel in the sum, while the product becomes a sum of squares, always giving a positive constant.

1. Example 1 — Roots 3 + 2i and 3 − 2i

Sum = (3 + 2i) + (3 − 2i) = 6. Product = (3 + 2i)(3 − 2i) = 9 − (2i)² = 9 − (−4) = 13. Equation: x² − 6x + 13 = 0.

2. Verify with the quadratic formula

x = (6 ± √(36 − 52))/2 = (6 ± √(−16))/2 = (6 ± 4i)/2 = 3 ± 2i ✓.

3. Example 2 — Purely imaginary roots: 4i and −4i

Sum = 4i + (−4i) = 0. Product = (4i)(−4i) = −16i² = −16(−1) = 16. Equation: x² + 0·x + 16 = 0 → x² + 16 = 0. Check: x² = −16, x = ±4i ✓.

Complex conjugate roots a ± bi always give the monic quadratic x² − 2ax + (a² + b²) = 0, where both coefficients are real.

Common Mistakes to Avoid

These four errors account for the majority of lost marks on problems that ask students to write the quadratic equation whose roots are given. Each mistake is easy to make under time pressure and just as easy to avoid once you know what to watch for.

1. Mistake 1 — Sign error in the factor form

The factor for root r is (x − r), not (x + r). For root −3, the factor is (x − (−3)) = (x + 3), not (x − 3). Writing (x − 3) instead produces roots of 3, not −3 — the sign of the constant term will be wrong.

2. Mistake 2 — Stopping at the factored form

After writing (x − r₁)(x − r₂) = 0, some students leave the answer in factored form. Unless the problem specifically asks for the factored form, expand fully to ax² + bx + c = 0.

3. Mistake 3 — Using sum directly without the minus sign

The Vieta template is x² − (sum)x + (product) = 0, not x² + (sum)x + (product) = 0. The coefficient of x is the negative of the sum. If the sum equals 7, the quadratic has −7x as its middle term, not +7x.

4. Mistake 4 — Not clearing fractions when required

If the problem asks for integer coefficients and the roots are fractions, multiply through after applying Vieta's formulas. For example, x² − (5/4)x + 3/8 = 0 must become 8x² − 10x + 3 = 0 by multiplying every term by 8.

Practice Problems with Full Solutions

Work through each problem before reading the solution. Use the factor method for Problems 1 and 2, Vieta's formulas for Problem 3, and your choice of method for Problems 4 and 5. These problems progress from straightforward integer roots to complex roots, matching the difficulty range on Algebra 2 and SAT practice tests.

1. Problem 1 — Roots 2 and 9

Sum = 2 + 9 = 11. Product = 2 × 9 = 18. Answer: x² − 11x + 18 = 0. Check: (x − 2)(x − 9) = x² − 9x − 2x + 18 = x² − 11x + 18 ✓.

2. Problem 2 — Roots −6 and −1

Sum = −6 + (−1) = −7. Product = (−6)(−1) = 6. Answer: x² − (−7)x + 6 = x² + 7x + 6 = 0. Check: (x + 6)(x + 1) = x² + x + 6x + 6 = x² + 7x + 6 ✓.

3. Problem 3 — Roots 1/3 and 2

Sum = 1/3 + 2 = 7/3. Product = (1/3)(2) = 2/3. Base equation: x² − (7/3)x + 2/3 = 0. Multiply by 3: 3x² − 7x + 2 = 0. Check: (3x − 1)(x − 2) = 3x² − 6x − x + 2 = 3x² − 7x + 2 ✓.

4. Problem 4 — Roots 1 + √2 and 1 − √2

Sum = (1 + √2) + (1 − √2) = 2. Product = (1 + √2)(1 − √2) = 1 − 2 = −1. Answer: x² − 2x − 1 = 0. Check via quadratic formula: x = (2 ± √(4 + 4))/2 = (2 ± 2√2)/2 = 1 ± √2 ✓.

5. Problem 5 — Roots 5 + i and 5 − i

Sum = 10. Product = (5 + i)(5 − i) = 25 − i² = 25 + 1 = 26. Answer: x² − 10x + 26 = 0. Check: discriminant = 100 − 104 = −4, roots = (10 ± 2i)/2 = 5 ± i ✓.

Quick self-check: substitute each root back into your equation. If both produce zero, the equation is correct.

Frequently Asked Questions

These questions come up regularly when students first learn to write the quadratic equation whose roots are specified. The answers address the most common points of confusion, from multiple valid answers to repeated roots and decimal inputs.

1. Can there be more than one correct quadratic for the same pair of roots?

Yes. If x² − 8x + 15 = 0 is one answer, then 2x² − 16x + 30 = 0 and 5x² − 40x + 75 = 0 are also correct — any nonzero scalar multiple works. Problems that want a unique answer typically specify 'monic form' (leading coefficient 1) or 'integer coefficients with GCF 1'.

2. What if both roots are the same (a repeated root)?

A repeated root r means r₁ = r₂ = r. The equation is (x − r)² = 0, which expands to x² − 2rx + r² = 0. For a repeated root of 4: (x − 4)² = x² − 8x + 16 = 0.

3. How do I handle decimal roots?

Apply Vieta's formulas the same way. For roots 0.5 and 1.5: sum = 2.0, product = 0.75. Equation: x² − 2x + 0.75 = 0. Multiply by 4 for integer coefficients: 4x² − 8x + 3 = 0. Verify: (4x − 2)(x − 1.5) → hmm, easier check: quadratic formula gives (8 ± √(64−48))/8 = (8 ± 4)/8 = 1.5 or 0.5 ✓.

4. Does the order of the roots matter?

No. (x − r₁)(x − r₂) and (x − r₂)(x − r₁) produce the same expansion by the commutative property of multiplication. List the roots in any order — the equation is identical.

5. What if only one root is given?

One root alone is not enough to define a unique quadratic unless you have extra information such as the sum or product, or the root is irrational/complex (in which case its conjugate is automatically the second root). For example, if you are told one root is 3 + √7, the other must be 3 − √7, giving sum = 6 and product = 9 − 7 = 2, so the equation is x² − 6x + 2 = 0.

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