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Functions, EXT1 F2 EQ-Bank 2

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)

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\(\text{Roots}\ = -1, 4, 5\)

Show Worked Solution

\(x^3-8 x^2+11x+20=0\)

\(\text{Sum of roots}\ =-\dfrac{b}{a}:\)

\(\alpha+\beta+(\alpha + \beta)\) \(=8\)  
\(2\alpha + 2\beta\) \(=8\)  
\(\beta\) \(=4-\alpha\ \ …\ (1)\)  

 

\(\text{Product of roots}\ =-\dfrac{d}{a}:\)

\(\alpha \beta(\alpha + \beta)=-20\ \ …\ (2) \)

\(\text{Substitute (1) into (2):}\)

\(\alpha(4-\alpha)(4)\) \(=-20\)  
\(16\alpha-4\alpha^{2}\) \(=-20\)  
\(\alpha^{2}-4\alpha-5\) \(=0\)  
\((\alpha-5)(\alpha+1)\) \(=0\)  

\(\alpha = 5\ \ \text{or}\ -1\)

\(\beta = 5\ \ \text{or}\ -1\ \ \text{(using (1))}\)

\(\alpha+ \beta = 5-1=4\)

\(\therefore \text{Roots}\ = -1, 4, 5\)

Functions, EXT1 F2 2021 HSC 11h

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)

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`1/2`

Show Worked Solution

`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`

Trigonometry, EXT1 T3 2020 HSC 14b

  1. Show that  `sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`.  (2 marks)

  2. By letting  `x = 4sin theta`  in the cubic equation  `x^3-12x + 8 = 0`.

     

    Show that  `sin (3theta) = 1/2`. (2 marks)

  3. Prove that  `sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`.  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution

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`
♦♦♦ Mean mark (iii) 21%.

 

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`

Functions, EXT1 F2 SM-Bank 5

The polynomial  `P(x) = x^3 + px^2 + qx + r`  has roots  `sqrtk`, `−sqrtk`  and  `alpha`.

  1.  Explain why  `alpha + p = 0`.  (1 mark)

  2.  Show that  `kalpha = r`.  (1 mark)

  3.  Show that  `pq = r`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution

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`

Functions, EXT1 F2 SM-Bank 2

The polynomial  `P(x) = x^3 - 2x^2 + kx + 24`  has roots  `alpha, beta, gamma`.

  1. Find the value of  `alpha + beta + gamma`.  (1 mark)

  2. Find the value of  `alphabetagamma`.  (1 mark)

  3. It is known that two of the roots are equal in magnitude but opposite in sign.

     

    Find the third root and hence find the value of `k`.  (2 marks)

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  1. `2`
  2. `−24`
  3. `−12`
Show Worked Solution

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`

Functions, EXT1 F2 SM-Bank 1

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`

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`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`

Functions, EXT1 F2 2019 HSC 7 MC

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`

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`C`

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`text(Roots:)\ \ alpha,\ beta,\ -1`

Mean mark 51%.

`alpha beta(-1) = -d/a = -q/q = -1`

`- alpha beta` `= -1`
`:. beta` `= 1/alpha`

 
`=>  C`

Functions, EXT1′ F2 2018 HSC 3 MC

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`?

  1. `x^3 - 5x^2 - 2x + 1 = 0`
  2. `x^3 - 5x^2 - 2x - 1 = 0`
  3. `x^3 + 5x^2 + 2x + 1 = 0`
  4. `x^3 + 5x^2 - 2x + 1 = 0`
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`text(D)`

Show Worked Solution

`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`

Functions, EXT1′ F2 2017 HSC 5 MC

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`?

  1. `−10`
  2. `−6`
  3. `−2`
  4. `0`
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`B`

Show Worked Solution

`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`

Functions, EXT1′ F2 2017 HSC 13b

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)}`

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`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.)`

Functions, EXT1′ F2 2007 HSC 3b

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`

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`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`

Functions, EXT1 F2 2006 HSC 4a

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.`

  1. Find the value of  `r.`  (1 mark)

  2. Find the value of  `s + t.`  (2 marks)

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  1. `-1`
  2. `0`
Show Worked Solution

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`

Functions, EXT1 F2 2008 HSC 2c

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`

Show Worked Solution

`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)`

 

MARKER’S COMMENT: Many students displayed significant inefficiencies in solving simultaneous equations.
`- 16/a – 1` `= -20/beta`
`-16 – a` `= -20`
`a` `= 4`

 
`text(Substitute)\ \ a = 4\ \ text(into)\ (1)`

`:. beta` `= – 16/4 – 1`
  `= -5`

Functions, EXT1 F2 2014 HSC 5 MC

Which group of three numbers could be the roots of the polynomial equation  `x^3 + ax^2 − 41x + 42 = 0`?

  1. `2, 3, 7`
  2. `1, −6, 7`
  3. `−1, −2, 21`
  4. `−1, −3, −14`
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`B`

Show Worked Solution

`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`

Functions, EXT1 F2 2012 HSC 3 MC

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`?

  1. `x^3 + 2x^2 + 3x + 1 = 0`
  2. `x^3 + 2x^2 + 3x- 1 = 0`
  3. `x^3- 2x^2 + 3x + 1 = 0`
  4. `x^3- 2x^2 + 3x- 1 = 0`
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`B`

Show Worked Solution

`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`