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It is given that \(\tan \alpha, \tan \beta\) and \(\tan \gamma\) are the three real roots of the polynomial equation \(x^3+b x^2+c x-1+b+c=0\), where \(b\) and \(c\) are real numbers and \(c \neq 1\).
Find the smallest positive value of \(\alpha+\beta+\gamma\). (3 marks)
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\(\alpha+\beta+\gamma=\dfrac{3 \pi}{4}\)
\(x^3+b x^2+c x-1+b+c=0\)
\(\text{Roots:} \ \tan \alpha, \tan \beta, \tan \gamma\)
\(\tan \alpha+\tan \beta+\tan \gamma=-\dfrac{b}{a}=-b\)
\(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma=c\)
\(\tan \alpha \cdot \tan \beta \cdot \tan \gamma=-\dfrac{d}{a}=1-b-c\)
\(\text { Find smallest +ve value of} \ \ \alpha+\beta+\gamma:\)
| \(\tan (\alpha+\beta+\gamma)\) | \(=\dfrac{\tan (\alpha+\beta)+\tan \gamma}{1-\tan (\alpha+\beta) \tan \gamma}\) |
| \(=\dfrac{\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}+\tan \gamma}{1-\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \times \tan \gamma}\) | |
| \(=\dfrac{\tan \alpha+\tan \beta+\tan \gamma(1-\tan \alpha \cdot \tan \beta)}{1-\tan \alpha \cdot \tan \beta-(\tan \alpha+\tan \beta) \tan \gamma}\) | |
| \(=\dfrac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \cdot \tan \beta \cdot \tan \gamma}{1-(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma)}\) | |
| \(=\dfrac{-b-(1-b-c)}{1-c}\) | |
| \(=\dfrac{-1+c}{1-c}\) | |
| \(=-1\) |
\(\therefore \ \text{Smallest +ve value of} \ \ \alpha+\beta+\gamma=\dfrac{3 \pi}{4}\)
The roots of \(2 x^3+6 x^2+x-1=0\) are \(\alpha, \beta\) and \(\gamma\).
What is the value of \(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}\) ? (2 marks)
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\(\dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=-6\)
\(2 x^3+6 x^2+x-1=0\)
\(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{\alpha+\beta+\gamma}{\alpha \beta \gamma}\)
\(\alpha+\beta+\gamma=-\dfrac{b}{a}=-3\)
\(\alpha \beta \gamma=-\dfrac{d}{a}=\dfrac{1}{2}\)
\(\therefore \dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{-3}{\frac{1}{2}}=-6\)
The polynomial \(P(x)=x^3+2 x^2-19 x-20\) has three distinct roots, where one root is the sum of the other two roots.
Find the values of the three roots. (3 marks)
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\(\text{Roots: \(-5,-1\) and \(4\).}\)
\(P(x)=x^3+2 x^2-19 x-20\)
\(\text{Let roots}=\alpha, \beta, \alpha+\beta\)
| \(\alpha+\beta+(\alpha+\beta)\) | \(=-\dfrac{b}{a}=-2\) |
| \(2 \alpha+2 \beta\) | \(=-2\) |
| \(\alpha\) | \(=-1-\beta\ \ldots\ (1)\) |
| \(\alpha \times \beta \times(\alpha+\beta)\) | \(=-\dfrac{d}{a}=20\) |
| \(\alpha^2 \beta+\alpha \beta^2\) | \(=20\ \ldots\ (2)\) |
\(\text{Substitute (1) into (2):}\)
| \((-1-\beta)^2 \beta+(-1-\beta) \beta^2\) | \(=20\) |
| \(\left(1+2 \beta+\beta^2\right) \beta+\left(-\beta^2-\beta^3\right)\) | \(=20\) |
| \(\beta+2 \beta^2+\beta^3-\beta^2-\beta^3\) | \(=20\) |
| \(\beta^2+\beta-20\) | \(=0\) |
| \((\beta+5)(\beta-4)\) | \(=0\) |
\(\Rightarrow \ \beta=-5 \ \ \text {or} \ \ 4\)
\(\text{If} \ \ \beta=-5, \alpha=4 \ \ \text {and if} \ \ \beta=4, \alpha=-5\)
\(\therefore \ \text{Roots: \(-5, -1\) and \(4\).}\)
The polynomial \(x^{3} + 2x^{2}-5x-6\) has zeros \(-1, -3\) and \(\alpha\).
What is the value of \(\alpha\)?
\(B\)
\(\alpha \beta \gamma = -\dfrac{\text{d}}{\text{a}} = 6\)
| \(-1 \times -3 \times \alpha\) | \(=6\) | |
| \(\alpha\) | \(=2\) |
\(\Rightarrow B\)
In the equation \(x^3-8 x^2+11x+20=0\) one of the roots is equal to the sum of the other two roots. Find the value of the three roots. (3 marks) --- 7 WORK AREA LINES (style=lined) --- \(\text{Roots}\ = -1, 4, 5\) \(x^3-8 x^2+11x+20=0\) \(\text{Sum of roots}\ =-\dfrac{b}{a}:\) \(\text{Product of roots}\ =-\dfrac{d}{a}:\) \(\alpha \beta(\alpha + \beta)=-20\ \ …\ (2) \) \(\text{Substitute (1) into (2):}\)
\(\alpha = 5\ \ \text{or}\ -1\) \(\beta = 5\ \ \text{or}\ -1\ \ \text{(using (1))}\) \(\alpha+ \beta = 5-1=4\) \(\therefore \text{Roots}\ = -1, 4, 5\)
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Show Worked Solution
\(\alpha+\beta+(\alpha + \beta)\)
\(=8\)
\(2\alpha + 2\beta\)
\(=8\)
\(\beta\)
\(=4-\alpha\ \ …\ (1)\)
\(\alpha(4-\alpha)(4)\)
\(=-20\)
\(16\alpha-4\alpha^{2}\)
\(=-20\)
\(\alpha^{2}-4\alpha-5\)
\(=0\)
\((\alpha-5)(\alpha+1)\)
\(=0\)
The roots of `x^4 - 3x + 6 = 0` are `alpha, beta, gamma` and `delta`.
What is the value of `1/alpha + 1/beta + 1/gamma + 1/delta`? (2 marks)
`1/2`
`alpha beta gamma delta = 6/1 = 6`
`alpha beta gamma + beta gamma delta + gamma delta alpha + delta alpha beta = – ((-3))/1 = 3`
| `1/alpha + 1/beta + 1/gamma + 1/delta` | `= (beta gamma delta + alpha gamma delta + beta gamma delta + alpha beta gamma)/(alpha beta gamma delta)` |
| `= 3/6` | |
| `= 1/2` |
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Show that `sin (3theta) = 1/2`. (2 marks)
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i. `text(Prove:)\ \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`
| `text(LHS)` | `= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)` |
| `= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1 – 2sin^2theta))` | |
| `= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)` | |
| `= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)` | |
| `= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta` | |
| `= 0` |
ii. `text(Show)\ \ sin(3theta) = 1/2`
`text{Using part (i):}`
| `(sin(3theta))/4` | `= 3/4 sintheta-sin^3 theta` |
| `sin(3theta)` | `= 3sintheta-4sin^3theta\ …\ (1)` |
`x^3-12x + 8 = 0`
`text(Let)\ \ x = 4 sin theta`
| `(4sintheta)^3-12(4sintheta) + 8` | `= 0` |
| `64sin^3theta-48sintheta` | `= 0` |
| `−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}` | `= −8` |
| `−16 sin(3theta)` | `= −8` |
| `sin(3theta)` | `= 1/2` |
iii. `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`
`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`
`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`
`text(When)\ \ sin3theta = 1/2,`
| `3theta` | `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …` |
| `theta` | `= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …` |
`:.\ text(Solutions)`
`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`
`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`
`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`
`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`
`α + β + γ = −b/a = 0`
`αβ + βγ + αγ = c/a = −12`
| `(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2` | `= (α + β + γ)^2 – 2(αβ + βγ + αγ)` |
| `16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)` | `= 0-2(−12)` |
| `:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18` | `= 24/16=3/2` |
The polynomial `P(x) = x^3 + px^2 + qx + r` has roots `sqrtk`, `−sqrtk` and `alpha`.
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i. `text(Sum of roots:)`
| `sqrtk – sqrtk + alpha` | `= −b/a` |
| `alpha` | `= −p` |
| `alpha + p` | `= 0` |
ii. `text(Product of roots:)`
| `sqrtk xx −sqrtk xx alpha` | `= −d/a` |
| `−kalpha` | `= −r` |
| `:.kalpha` | `= r` |
| iii. | `sqrtk(−sqrtk) + sqrtk alpha – sqrtk alpha` | `= c/d` |
| `−k` | `= q` |
`text(Substitute)\ \ k = −q\ \ text(into part (ii)):`
| `−qalpha` | `= r` |
| `−q xx – p` | `= r\ \ \ text{(using part(i))}` |
| `:. pq` | `= r` |
The polynomial `P(x) = x^3 - 2x^2 + kx + 24` has roots `alpha, beta, gamma`.
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Find the third root and hence find the value of `k`. (2 marks)
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i. `alpha + beta + gamma = −b/a = 2`
ii. `alphabetagamma = −d/a = −24`
iii. `text(Let roots be)\ \ alpha, −alpha, \ beta:`
| `alpha – alpha + beta` | `= 2` |
| `beta` | `= 2` |
`text(Substitute)\ \ beta = 2\ \ text(into equation:)`
| `2^3 – 2 ·2^2 + 2k + 24` | `= 0` |
| `2k` | `= −24` |
| `k` | `= −12` |
The equation `2x^3 + x^2 - kx + 6 = 0` has 2 roots which are reciprocals of each other.
Find the value of `k`. (2 marks)
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`13`
`text(Let roots be)\ \ alpha, \ 1/alpha, \ beta:`COMMENT: Substituting `beta=-3` into the equation solves this question efficiently.
| `alpha · 1/alpha · beta` | `= −d/a` |
| `beta` | `= −6/2` |
| `= −3` |
`text(Substitute)\ \ beta = −3\ \ text(into equation:)`
| `2(−3)^3 + (−3)^2 – (−3)k + 6` | `= 0` |
| `−54 + 9 + 3k + 6` | `= 0` |
| `3k` | `= 39` |
| `k` | `= 13` |
Let `P(x) = qx^3 + rx^2 + rx + q` where `q` and `r` are constants, `q != 0`. One of the zeros of `P(x)` is `-1`.
Given that ` alpha` is a zero of `P(x),\ alpha != -1`, which of the following is also a zero?
A. `-1/alpha`
B. `-q/alpha`
C. `1/alpha`
D. `q/alpha`
`C`
`text(Roots:)\ \ alpha,\ beta,\ -1`Mean mark 51%.
`alpha beta(-1) = -d/a = -q/q = -1`
| `- alpha beta` | `= -1` |
| `:. beta` | `= 1/alpha` |
`=> C`
The cubic equation `x^3 + 2x^2 + 5x - 1 = 0` has roots, `alpha`, `beta` and `gamma`.
Which cubic equation has roots `(−1)/alpha, (−1)/beta, (−1)/gamma`?
`text(D)`
`alpha beta gamma = 1`
`alpha beta + beta gamma + gamma alpha = 5`
`alpha + beta + gamma = −2`
`text(If roots are)\ \ (−1)/alpha, (−1)/beta, (−1)/gamma :`
`(−1)/alpha · (−1)/beta · (−1)/gamma = (−1)/(alphabetagamma) = −1`
`=> d = 1`
`1/(alphabeta) + 1/(betagamma) + 1/(gammaalpha) = ((alpha + beta + gamma))/(alphabetagamma) = −2`
`=>\ c = −2`
`(−1)/alpha – (−1)/beta – (−1)/gamma = (−(alphabeta + betagamma + gammaalpha))/(alphabetagamma) = −5`
`=> b = 5`
`:.\ text(Equation is)\ \ x^3 + 5x^2 – 2x + 1 = 0`
`=>D`
The polynomial `p(x) = x^3 - 2x + 2` has roots `alpha`, `beta` and `gamma`.
What is the value of `alpha^3 + beta^3 + gamma^3`?
`B`
`text(S)text(ince)\ \ alpha, beta, gamma\ text(are roots,)`
| `p(alpha)` | `= alpha^3 – 2alpha + 2 = 0` | `…\ (1)` |
| `p(beta)` | `= beta^3 – 2beta + 2 = 0` | `…\ (2)` |
| `p(gamma)` | `= gamma^3 – 2gamma + 2 = 0` | `…\ (3)` |
`text(Add:)\ (1) + (2) + (3)`
`alpha^3 + beta^3 + gamma^3 – 2(alpha + beta + gamma) + 6 = 0`
`:. alpha^3 + beta^3 + gamma^3=-6\ \ \ \ text{(note:}\ \ alpha + beta + gamma =- b/a=0 text{)}`
`=> B`
Let `a, b` and `c` be real numbers. Suppose that `P(x) = x^4 + ax^3 + bx^2 + cx + 1` has roots `alpha, 1/alpha, beta, 1/beta,` where `alpha > 0 and beta > 0`.
Prove that `a = c`. (2 marks)
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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Sum of roots) = -b/a:`
`-a = alpha + 1/alpha + beta + 1/beta`
`text(Sum of products of 3 roots) = -d/a:`
| `- c` | `= alpha · 1/alpha · beta + alpha · 1/alpha · 1/beta + alpha · beta · 1/beta + 1/alpha · beta · 1/beta` |
| `= beta + 1/beta + alpha + 1/alpha` | |
| `= -a` |
`:. a = c\ text(… as required.)`
The equation `2x^3 − 3x^2 − 5x − 1 = 0` has roots `α`, `β` and `γ`.
What is the value of `1/(α^3β^3γ^3)`?
`C`
| `αβγ` | `=(-d)/a= 1/2` |
| `:.α^3β^3γ^3` | `=(1/2)^3=1/8` |
| `:.1/(α^3β^3γ^3)` | `=8` |
`=>C`
The zeros of `x^3 - 5x + 3` are `alpha, beta` and `gamma.`
Find a cubic polynomial with integer coefficients whose zeros are `2 alpha, 2 beta` and `2 gamma.` (2 marks)
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`x^3 – 20x + 24`
`text(Solution 1)`
| `α+β+γ` | `=0` |
| `:.2α+2β+2γ` | `=2(α+β+γ)` |
| `=0` | |
| `αβ+βγ+γα` | `=-5` |
| `:.2α2β+2β2γ+2γ2α` | `=4(αβ+βγ+γα)` |
| `=-20` | |
| `αβγ` | `=-3` |
| `:.2α2β2γ` | `=8αβγ` |
| `=-24` |
`:.\ text(Polynomial is)\ \ x^3 – 20x + 24`
`text(Solution 2)`
`P(x) = x^3 – 5x + 3`
`text(New zeros are)\ \ 2 alpha, 2 beta and 2 gamma`
| `:.H(x)` | `=(x/2)^3 – 5(x/2) + 3` |
| `=x^3/8-(5x)/2+3` |
`:.\ text(New Polynomial with integer coefficients is)`
`x^3 – 20x + 24`
The cubic polynomial `P(x) = x^3 + rx^2 + sx + t`. where `r, \ s` and `t` are real numbers, has three real zeros, `1, alpha` and `-alpha.`
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i. `P(x) = x^3 + rx^2 + sx + t`
`text(Roots are)\ \ 1, alpha, -alpha`
| `1 + alpha + -alpha` | `= – b/a` |
| `1` | `= – r/1` |
| `:.\ r` | `= -1` |
ii. `P(x) = x^3 – x^2 + sx + t`
`P(1) = 0`
| `0` | `= 1 – 1 + (s xx 1) + t` |
| `0` | `= s + t` |
`:.\ text(Value of)\ \ s + t = 0`
The polynomial `p(x)` is given by `p(x) = ax^3 + 16x^2 + cx - 120`, where `a` and `c` are constants.
The three zeros of `p(x)` are `– 2`, `3` and `beta`.
Find the value of `beta`. (3 marks)
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`- 5`
`p(x) = ax^3 + 16x^2 + cx – 120`
`text(Roots:)\ \ – 2, \ 3, \ beta`
| `-2 + 3 + beta` | `= -B/A` |
| `beta + 1` | `= -16/a` |
| `beta` | `= -16/a – 1\ \ \ \ \ …\ (1)` |
| `-2 xx 3 xx beta` | `= -D/A` |
| `-6 beta` | `= 120/a` |
| `beta` | `= -20/a\ \ \ \ \ …\ (2)` |
| `- 16/a – 1` | `= -20/beta` |
| `-16 – a` | `= -20` |
| `a` | `= 4` |
`text(Substitute)\ \ a = 4\ \ text(into)\ (1)`
| `:. beta` | `= – 16/4 – 1` |
| `= -5` |
Which group of three numbers could be the roots of the polynomial equation `x^3 + ax^2 − 41x + 42 = 0`?
`B`
`text(Let roots be)\ alpha, beta, gamma`
`alpha beta gamma = -d/a = -42`
`:.\ text(Cannot be)\ A\ text(or)\ C`
`alpha beta + beta gamma + gamma alpha = c/a = -41`
`:.\ text(Cannot be D)`
`=> B`
The polynomial equation `2x^3- 3x^2- 11x + 7 = 0` has roots `alpha`, `beta` and `gamma`.
Find `alpha beta gamma`. (1 mark)
`-7/2`
`P(x) = 2x^3- 3x^2- 11x + 7 = 0`
`alpha beta gamma = -d/a = -7/2`
A polynomial equation has roots `alpha`, `beta`, `gamma` where
`alpha + beta + gamma = -2`, `alphabeta + alphagamma + betagamma = 3`, `alphabetagamma = 1`.
Which polynomial equation has the roots `alpha`, `beta`, and `gamma`?
`B`
`text(Using)\ \ ax^3 + bx^2 + cx + d = 0`
`text(By elimination)`
`alpha beta gamma = -d/a = 1`
`:.\ text(Cannot be)\ A\ text(or)\ C\ \ (text(where)\ alphabetagamma = -1 text{)}`
`alpha + beta + gamma = -b/a = -2`
`:.\ text(Cannot be)\ D\ \ (text(where)\ alpha + beta + gamma = 2 text{)}`
`=> B`