Forming a New Quadratic Equation From Its Roots

Suppose you know that alpha and beta are the roots of a quadratic. A question then asks for an equation whose roots are their reciprocals, their squares, or the same numbers increased by two. Do you have to solve the first equation, find both roots, and start again?

Usually, no. Think of the original roots as a pair of travellers. You can change both in the requested way while keeping track of just two things: their sum and their product. Those two numbers are enough to build the new quadratic.

We will work through reciprocal, square, shift, and scale questions, including cases where the original roots are awkward to calculate. The method remains the same throughout.

The one idea that starts every question

If the roots of ax² + bx + c = 0, where a is not zero, are alpha and beta, then:

S = alpha + beta = -b/a
P = alpha x beta = c/a

Why? The original expression can be written as a(x - alpha)(x - beta). Expanding gives ax² - a(alpha + beta)x + a(alpha x beta), so matching coefficients gives the two relations. You do not need to find alpha and beta separately.

Now call the new roots u and v. Once you have their sum Snew = u + v and product Pnew = uv, write:

x² - Snew x + Pnew = 0

If fractions appear, multiply the whole equation by a common denominator. Multiplying every term by the same nonzero number changes its appearance, not its roots.

As a running example, let the original equation be 2x² - 7x + 3 = 0. Its roots are 3 and 1/2, but the sum-and-product method does not need that information:

S = 7/2
P = 3/2

We will use this same starting point to see exactly what each change does.

Reciprocal roots: 1/alpha and 1/beta

The two new roots are 1/alpha and 1/beta. Add and multiply them before writing an equation:

Snew = 1/alpha + 1/beta
     = (alpha + beta)/(alpha x beta)
     = S/P

Pnew = 1/(alpha x beta) = 1/P

For 2x² - 7x + 3 = 0, this gives Snew = (7/2)/(3/2) = 7/3 and Pnew = 1/(3/2) = 2/3.

x² - (7/3)x + 2/3 = 0

Multiply every term by 3:
3x² - 7x + 2 = 0

Check with the known roots if you like: reciprocals of 3 and 1/2 are 1/3 and 2. They add to 7/3 and multiply to 2/3.

There is a handy coefficient shortcut. From ax² + bx + c = 0, the equation with reciprocal roots is:

cx² + bx + a = 0

The outer coefficients exchange places; the middle one stays put. This also follows by replacing the old variable with 1/x and multiplying through by x². The sum-and-product steps explain why the shortcut works.

Squared roots: alpha² and beta²

The product is easy: alpha² x beta² = P². The sum needs one identity:

alpha² + beta²
= (alpha + beta)² - 2(alpha x beta)
= S² - 2P

For our running example, the new sum is (7/2)² - 2(3/2) = 49/4 - 3 = 37/4. The new product is (3/2)² = 9/4.

x² - (37/4)x + 9/4 = 0

Multiply every term by 4:
4x² - 37x + 9 = 0

The direct check is reassuring: 3² = 9 and (1/2)² = 1/4; their sum is 37/4 and their product is 9/4.

If you want a shortcut in the original coefficients, substitute S = -b/a and P = c/a into the new equation and multiply by a²:

a²x² - (b² - 2ac)x + c² = 0

The new roots might coincide even when the old ones do not. For example, the roots of x² - 1 = 0 are 1 and -1. Squaring both gives 1 and 1, so the new equation is (x - 1)² = 0. A repeated root still counts twice when you form the quadratic.

Shifted roots: alpha + k and beta + k

Adding the same number k to both roots changes the sum by 2k, because there are two roots. Their product needs brackets:

Snew = (alpha + k) + (beta + k)
     = S + 2k

Pnew = (alpha + k)(beta + k)
     = P + kS + k²

Take k = 2 with 2x² - 7x + 3 = 0. Then:

Snew = 7/2 + 4 = 15/2
Pnew = 3/2 + 2(7/2) + 4 = 25/2

x² - (15/2)x + 25/2 = 0
2x² - 15x + 25 = 0

The original roots 3 and 1/2 become 5 and 5/2, which confirms the answer.

There is a second route worth understanding. Let the new variable be x = old root + k, so the old root is x - k. Substitute x - k into the original equation. For the same example:

2(x - 2)² - 7(x - 2) + 3 = 0
2x² - 15x + 25 = 0

That minus sign is the classic trap: new roots alpha + k require substituting x - k, not x + k, in the old expression. In contrast, if the requested new roots are alpha - k and beta - k, substitute x + k.

Scaled roots: m alpha and m beta

Multiplying each root by m multiplies the sum by m and the product by m²:

Snew = mS
Pnew = m²P

If both roots of 2x² - 7x + 3 = 0 are doubled, their new sum is 2(7/2) = 7, and their new product is 4(3/2) = 6. The equation is x² - 7x + 6 = 0, with roots 6 and 1.

The product does not merely double: it contains two multiplied roots. For m = -1, this also gives the equation for the negatives of the old roots. Notice that m = 0 turns both new roots into zero, so the resulting equation is x² = 0.

When the original roots refuse to factor nicely

The method is most valuable when solving the original equation would create radicals. Consider x² - 4x + 1 = 0. We know S = 4 and P = 1, without needing to write its roots.

• For reciprocal roots, Snew = S/P = 4 and Pnew = 1/P = 1, so the new equation is again x² - 4x + 1 = 0. A pair with product 1 swaps places when both roots are reciprocated.
• For squared roots, Snew = S² - 2P = 16 - 2 = 14 and Pnew = P² = 1, so the new equation is x² - 14x + 1 = 0.
• For roots each decreased by 3, use k = -3. Then Snew = 4 - 6 = -2 and Pnew = 1 - 12 + 9 = -2, giving x² + 2x - 2 = 0.

The answers are exact. Finding the original roots first would add work without giving you anything the question needs.

A two-step change: reciprocate after shifting

Some questions combine operations. The order matters. Suppose alpha and beta solve x² - 5x + 6 = 0, and the required roots are 1/(alpha + 1) and 1/(beta + 1).

First shift both roots by 1. The original sum is 5 and product is 6, so the shifted roots have sum 5 + 2 = 7 and product 6 + 5 + 1 = 12.

Then take reciprocals of those shifted roots:

Snew = 7/12
Pnew = 1/12

x² - (7/12)x + 1/12 = 0
12x² - 7x + 1 = 0

The old roots happen to be 2 and 3. After adding 1, they are 3 and 4, and their reciprocals are 1/3 and 1/4. That independently checks the answer.

You may take these reciprocals only when neither shifted root is zero. For 1/(alpha + k) and 1/(beta + k), check that P + kS + k² is not zero before dividing by it. If one old root equals -k, the requested pair is not defined.

A quick decision guide

You can keep the whole lesson on a small revision card. Start from S = -b/a, P = c/a, and use x² - Snew x + Pnew = 0 after the change.

New roots	New sum	New product
1/alpha, 1/beta	S/P	1/P
alpha², beta²	S² - 2P	P²
alpha + k, beta + k	S + 2k	P + kS + k²
m alpha, m beta	mS	m²P

For reciprocals, check P != 0. For a combined change, complete one row at a time and carry the new sum and product into the next row.

One further check catches many mistakes: substitute roots when they are simple. If not, compare the sum and product implied by your proposed answer with the values you calculated. In Ax² + Bx + C = 0, that final check is -B/A for the sum and C/A for the product. A different nonzero multiple of the same equation is equally valid.

Try these yourself

Pause before the answers and write S and P for each starting equation.

1. If x² - 5x + 6 = 0 has roots alpha and beta, form the equation with roots 1/alpha and 1/beta.
2. If 3x² - 4x + 1 = 0 has roots alpha and beta, form the equation with roots alpha² and beta².
3. If x² - 3x - 2 = 0 has roots alpha and beta, form the equation with roots alpha - 4 and beta - 4.
4. If x² + 2x - 8 = 0 has roots alpha and beta, form the equation with roots -alpha and -beta.
5. If x² - 1 = 0 has roots alpha and beta, form the equation with roots alpha² and beta².
6. If 2x² - 7x + 3 = 0 has roots alpha and beta, form the equation with roots 2alpha + 1 and 2beta + 1.

Answers with the key step

1. S = 5, P = 6. Reciprocal sum 5/6, product 1/6: 6x² - 5x + 1 = 0.
2. S = 4/3, P = 1/3. Squared sum 16/9 - 2/3 = 10/9, product 1/9: 9x² - 10x + 1 = 0.
3. S = 3, P = -2, k = -4. New sum -5, product -2 - 12 + 16 = 2: x² + 5x + 2 = 0.
4. S = -2, P = -8, m = -1. New sum 2, product -8: x² - 2x - 8 = 0.
5. S = 0, P = -1. Squared sum 0 - 2(-1) = 2, product 1: x² - 2x + 1 = 0. Both new roots equal 1.
6. S = 7/2, P = 3/2. Doubling gives sum 7 and product 6; adding 1 gives sum 9 and product 6 + 7 + 1 = 14: x² - 9x + 14 = 0.

If you missed a sign, go back to the new roots and rebuild their sum and product. That is much more helpful than trying to memorise a longer list of final coefficient formulas.

Sources for further study

• UCLA Math Circle: Vieta’s formula derives the sum and product of quadratic roots and the equation formed from them.
• Emory Mathematics: Quadratics and Vieta’s formulas explains why the coefficients depend on the symmetric sum and product of the roots.
• OpenStax College Algebra: Quadratic Equations revisits factored form, the zero-product property, and repeated roots.
• University of Washington Math Circle: Vieta’s formulas develops the broader root-and-coefficient relationship.

Frequently asked questions

Must I solve the original quadratic first?

No. Read its root sum -b/a and product c/a, transform those two values, and form x² - Snew x + Pnew = 0. Solving first is useful only as an optional check when the roots are simple.

Why is the sign before the new sum negative?

Because (x - u)(x - v) = x² - (u + v)x + uv. The product is the constant term; the negative of the sum is the coefficient of x in the equation with leading coefficient 1.

Can I swap a and c to find reciprocal roots every time?

Only when c is not zero. If c = 0, one original root is zero and 1/0 is undefined. When both roots are nonzero, cx² + bx + a = 0 is the reciprocal-root equation.

Why does a shift by k add 2k to the sum?

Both roots shift. (alpha + k) + (beta + k) = alpha + beta + 2k. Their product is (alpha + k)(beta + k), so it also includes the cross term k(alpha + beta).

Why do I substitute x - k for roots alpha + k?

If a new root is x = alpha + k, the corresponding old root is alpha = x - k. Replacing the old variable by x - k therefore makes the old equation vanish at the desired new root.

What if the two squared roots become equal?

Keep both copies of the repeated root. If alpha and beta are 1 and -1, both squares are 1, and the requested quadratic is (x - 1)² = 0.

Is my answer wrong if every coefficient differs from the answer key?

Not necessarily. x² - 5x + 6 = 0 and 2x² - 10x + 12 = 0 have exactly the same roots. Divide out any common nonzero factor and compare the equations again.

Can a combined shift and reciprocal be done in either order?

No. 1/(alpha + 1) is generally different from 1/alpha + 1. Follow the brackets in the requested roots: shift first, check that neither shifted root is zero, and only then reciprocate.