# 2020 Normanhurst Ext 1 Trial Paper Help (1 Viewer)

#### csi

##### Member
Hi,

Can I please have a hand on these two questions:

Consider the polynomial P(x) = x^3+x^2+cx-10. It is known that two of its zeros are equal in magnitude but opposite in sign. What is the value of c?
(a) root 10
(b) 10
(c) -10
(d) - root 10

Thanks!!

#### B1andB2

##### oui oui baguette
Let the roots be alpha, - alpha, beta

sum of roots

alpha + (- alpha) + beta = -1
So beta = -1

product of roots
-alpha^2 x beta = 10
Alpha ^2 = 10 (1)

sum two at a time
- alpha^2 = c (2)

sub 1 into 2

c = -10

for the other one I think the answer is 14

i would do 9C3/3! (Not completely certain)

Last edited:

#### csi

##### Member
Let the roots be alpha, - alpha, beta

sum of roots

alpha + (- alpha) + beta = -1
So beta = -1

product of roots
-alpha^2 x beta = 10
Alpha ^2 = 10 (1)

sum two at a time
- alpha^2 = c (2)

sub 1 into 2

c = -10

for the other one I’m pretty sure the answer is 14

i would do 9C3/3!
Thank you

#### idkkdi

##### Well-Known Member
Let the roots be alpha, - alpha, beta

sum of roots

alpha + (- alpha) + beta = -1
So beta = -1

product of roots
-alpha^2 x beta = 10
Alpha ^2 = 10 (1)

sum two at a time
- alpha^2 = c (2)

sub 1 into 2

c = -10

for the other one I think the answer is 14

i would do 9C3/3! (Not completely certain)
Thank you
9C3/3! seems kind of sketchy.
may be better to split it into |---|.
two diagonal combinations of points.
two vertical combinations of points.
5C3 for the horizontal line.

#### Qeru

##### Member
Hi,

Can I please have a hand on these two questions:

Consider the polynomial P(x) = x^3+x^2+cx-10. It is known that two of its zeros are equal in magnitude but opposite in sign. What is the value of c?
(a) root 10
(b) 10
(c) -10
(d) - root 10

View attachment 29844

Thanks!!
Another cool way of doing the poly q: sub $\bg_white \alpha, -\alpha$ into the original equation to get: $\bg_white (\alpha)^3+(\alpha)^2+c(\alpha)-10=0$ and $\bg_white (-\alpha)^3+(-\alpha)^2+c(-\alpha)-10=0$ Add these two to get:$\bg_white 2(\alpha)^2-20=0$ i.e. $\bg_white \alpha=\sqrt{10}$ WLOG. Making c the subject:
$\bg_white c=\frac{10-\alpha^2-\alpha^3}{\alpha}$

At this point you could just use your calculator (since its M.C) but working it out:
$\bg_white c=\frac{10-10-10\sqrt{10}}{\sqrt{10}}$
$\bg_white c=-10$

OR even better looking at B1B2's response, she found a root $\bg_white \beta=-1$ from here we can simply sub this into the equation to find c, $\bg_white (-1)^3+(-1)^2-c-10=0 \implies c=-10$

Last edited:

##### Member
Let the roots be alpha, - alpha, beta

sum of roots

alpha + (- alpha) + beta = -1
So beta = -1

product of roots
-alpha^2 x beta = 10
Alpha ^2 = 10 (1)

sum two at a time
- alpha^2 = c (2)

sub 1 into 2

c = -10

for the other one I think the answer is 14

i would do 9C3/3! (Not completely certain)
hmm for q10.. i dunno if it is 14 just because of the fact that I have tried like 10 different combinations of these 9 dots and a max of 8 combinations is all I found. Might be wrong but I'm just trying to help (I have no real proof - might be pigeonhole principle who knows)

csi

#### Qeru

##### Member
hmm for q10.. i dunno if it is 14 just because of the fact that I have tried like 10 different combinations of these 9 dots and a max of 8 combinations is all I found. Might be wrong but I'm just trying to help (I have no real proof - might be pigeonhole principle who knows)
Do what Idkddi did. 2 combinations diagonally, 2 combinations vertically. Then horizontally 5 dots need to choose 3 $\bg_white 5C3$ so in total:$\bg_white Ways=4+5C3=10+4=14$

#### csi

##### Member
Another cool way of doing the poly q: sub $\bg_white \alpha, -\alpha$ into the original equation to get: $\bg_white (\alpha)^3+(\alpha)^2+c(\alpha)-10=0$ and $\bg_white (-\alpha)^3+(-\alpha)^2+c(-\alpha)-10=0$ Add these two to get:$\bg_white 2(\alpha)^2-20=0$ i.e. $\bg_white \alpha=\sqrt{10}$ WLOG. Making c the subject:
$\bg_white c=\frac{10-\alpha^2-\alpha^3}{\alpha}$

At this point you could just use your calculator (since its M.C) but working it out:
$\bg_white c=\frac{10-10-10\sqrt{10}}{\sqrt{10}}$
$\bg_white c=-10$

OR even better looking at B1B2's response, she found a root $\bg_white \beta=-1$ from here we can simply sub this into the equation to find c, $\bg_white (-1)^3+(-1)^2-c-10=0 \implies c=-10$
Thanks