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Functions - Sum and product of roots (HL)

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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For a polynomial of degree nn in the form anxn+an−1xn−1+⋯+a0=0a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0, Vieta's formulas state that the sum of the roots is always −an−1an-\frac{a_{n-1}}{a_n}. This relationship connects the roots to the second-highest degree coefficient.

Graph of a quadratic function showing roots -1 and 3, where the sum equals 2, matching the coefficient ratio.
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For a cubic equation ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 with roots α,β,γ\alpha, \beta, \gamma, the 'sum of products taken two at a time' is given by αβ+βγ+γα=ca\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}. This relates to the linear term coefficient.

Graph of a cubic function x^3 - 3x showing three roots where the sum is 0.
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The product of roots for a polynomial depends on its degree nn. For a cubic (n=3n=3), the product αβγ=−da\alpha\beta\gamma = -\frac{d}{a}. Generally, for degree nn, the product is (−1)na0an(-1)^n \frac{a_0}{a_n}.

A table showing the alternating sign of the product of roots depending on whether the polynomial degree is even or odd.
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Algebraic identities are frequently used with Vieta's formulas to find sums of squares α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 or reciprocals 1α+1β+1γ\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}. This method avoids solving for individual roots.

Conceptual box representing the transformation of polynomial coefficients into root-based expressions using identities.

📐Formulae

Quadratic: α+β=−ba\text{Quadratic: } \alpha + \beta = -\frac{b}{a}

Quadratic: αβ=ca\text{Quadratic: } \alpha\beta = \frac{c}{a}

Cubic: α+β+γ=−ba\text{Cubic: } \alpha + \beta + \gamma = -\frac{b}{a}

Cubic: αβ+βγ+γα=ca\text{Cubic: } \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}

Cubic: αβγ=−da\text{Cubic: } \alpha\beta\gamma = -\frac{d}{a}

α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)

💡Examples

Problem 1:

Given the quadratic equation 2x2−8x+k=02x^2 - 8x + k = 0, where one root is three times the other root, find the value of kk.

Solution:

Let the roots be α\alpha and 3α3\alpha. Using Vieta's formulas:

  1. Sum of roots: α+3α=−−82=4\alpha + 3\alpha = -\frac{-8}{2} = 4. This gives 4α=44\alpha = 4, so α=1\alpha = 1.
  2. Product of roots: α(3α)=k2\alpha(3\alpha) = \frac{k}{2}. Substituting α=1\alpha = 1, we get 3(1)2=k23(1)^2 = \frac{k}{2}, which simplifies to 3=k23 = \frac{k}{2}.
  3. Therefore, k=6k = 6.

Explanation:

We identify the roots in terms of a single variable based on the given ratio, then apply the sum and product relationships to solve for the unknown coefficient.

Problem 2:

Find the sum of the squares of the roots for the cubic equation x3−5x2+7x−2=0x^3 - 5x^2 + 7x - 2 = 0.

Solution:

Let the roots be α,β,γ\alpha, \beta, \gamma. From the coefficients: Sum: α+β+γ=−−51=5\alpha + \beta + \gamma = -\frac{-5}{1} = 5. Sum of pairs: αβ+βγ+γα=71=7\alpha\beta + \beta\gamma + \gamma\alpha = \frac{7}{1} = 7. We need to find α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2. Using the identity: α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) α2+β2+γ2=(5)2−2(7)\alpha^2 + \beta^2 + \gamma^2 = (5)^2 - 2(7) α2+β2+γ2=25−14=11\alpha^2 + \beta^2 + \gamma^2 = 25 - 14 = 11.

Explanation:

By using the coefficients of the cubic, we determine the elementary symmetric sums and then apply the algebraic identity for the sum of squares.

Problem 3:

If the roots of x3+px2+qx+r=0x^3 + px^2 + qx + r = 0 are α,β,γ\alpha, \beta, \gamma, find the value of 1α+1β+1γ\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}.

Solution:

The expression 1α+1β+1γ\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} can be rewritten by finding a common denominator: 1α+1β+1γ=βγ+αγ+αβαβγ\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} From Vieta's formulas: Sum of pairs αβ+βγ+γα=q\alpha\beta + \beta\gamma + \gamma\alpha = q Product αβγ=−r\alpha\beta\gamma = -r Substituting these values: 1α+1β+1γ=q−r=−qr\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{q}{-r} = -\frac{q}{r}

Explanation:

We manipulate the required expression into a form involving the known sum and product relationships derived from the cubic polynomial coefficients.

Problem 4:

A cubic equation 2x3−6x2+4x−5=02x^3 - 6x^2 + 4x - 5 = 0 has roots α,β,γ\alpha, \beta, \gamma. Determine the value of (α+1)(β+1)(γ+1)(\alpha+1)(\beta+1)(\gamma+1).

Graph of the cubic function showing the value at x = -1, which helps determine the product of shifted roots.

Solution:

Let P(x)=2x3−6x2+4x−5P(x) = 2x^3 - 6x^2 + 4x - 5. The roots are α,β,γ\alpha, \beta, \gamma. We know P(x)=2(x−α)(x−β)(x−γ)P(x) = 2(x-\alpha)(x-\beta)(x-\gamma). We want to find (α+1)(β+1)(γ+1)(\alpha+1)(\beta+1)(\gamma+1). Note that P(−1)=2(−1−α)(−1−β)(−1−γ)=2[(−1)(1+α)(−1)(1+β)(−1)(1+γ)]=−2(1+α)(1+β)(1+γ)P(-1) = 2(-1-\alpha)(-1-\beta)(-1-\gamma) = 2[(-1)(1+\alpha)(-1)(1+\beta)(-1)(1+\gamma)] = -2(1+\alpha)(1+\beta)(1+\gamma). Calculate P(−1)=2(−1)3−6(−1)2+4(−1)−5=−2−6−4−5=−17P(-1) = 2(-1)^3 - 6(-1)^2 + 4(-1) - 5 = -2 - 6 - 4 - 5 = -17. Therefore, −2(α+1)(β+1)(γ+1)=−17-2(\alpha+1)(\beta+1)(\gamma+1) = -17. (α+1)(β+1)(γ+1)=−17−2=8.5(\alpha+1)(\beta+1)(\gamma+1) = \frac{-17}{-2} = 8.5.

Explanation:

This example uses the factor theorem. Instead of calculating αβγ+∑αβ+∑α+1\alpha\beta\gamma + \sum\alpha\beta + \sum\alpha + 1, we substitute x=−1x = -1 into the polynomial, which is equivalent to the scaled product of the shifted roots.

Problem 5:

For the quadratic equation x2−px+q=0x^2 - px + q = 0, the roots are α\alpha and β\beta. Find the value of pp and qq if α+β=5\alpha + \beta = 5 and α2+β2=13\alpha^2 + \beta^2 = 13.

Graph of x^2 - 5x + 6 showing roots at 2 and 3, which sum to 5 and have squares summing to 13.

Solution:

From Vieta's formulas: α+β=−(−p)1=p\alpha + \beta = \frac{-(-p)}{1} = p αβ=q1=q\alpha\beta = \frac{q}{1} = q Given p=5p = 5. Use the identity: α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta Substitute the known values: 13=(5)2−2q13 = (5)^2 - 2q 13=25−2q13 = 25 - 2q 2q=25−13=122q = 25 - 13 = 12 q=6q = 6 So, p=5p = 5 and q=6q = 6.

Explanation:

We use the relationship between coefficients and the sum/product of roots. The sum of squares identity allows us to find the product qq once the sum pp is known.