# Further maths help needed!!!Watch

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#1
Solve the equation 24x3+28x2-14x-3=0 given that the roots are a, a/r and ar

What I did was, I put the roots into brackets, so

(x-a)(x-a/r)(x-ar)

I expand this and put -a3 equal to -3, and get 1.44, although when I try to find r I get -0.049, -20.395 and -0.13 and -7.6

Can someone help me see where I have gone wrong?
0
4 weeks ago
#2
(Original post by Juliakinga)
Solve the equation 24x3+28x2-14x-3=0 given that the roots are a, a/r and ar

What I did was, I put the roots into brackets, so

(x-a)(x-a/r)(x-ar)

I expand this and put -a3 equal to -3, and get 1.44, although when I try to find r I get -0.049, -20.395 and -0.13 and -7.6

Can someone help me see where I have gone wrong?
0
4 weeks ago
#3
Did you multiply by r before you compared coefficients? Your expansion gives x^3 but your original equation is 24x^3.
0
#4
(Original post by Muttley79)
(X-a)(x-a/r)(x-ar)=
x3-arx2-ax2/r+a2x-ax2+a2rx+a2x/r-a2

-a3=-3
a=1.44

a2r+a2r2+a2=-14r
a2r+a2r2+a2+14r=0

2.0736r+2.0736r2+2.0736+14r=0
r=-0.13
r=-7.6

I think what I did wrong was I didn't multiply the second equation by 24 so that the x3 values are equal
Last edited by Juliakinga; 4 weeks ago
0
4 weeks ago
#5
(Original post by Juliakinga)
(X-a)(x-a/r)(x-ar)=
x3-arx2-ax2/r+a2x-ax2+a2rx+a2x/r-a2

I think what I did wrong was I didn't multiply the second equation by 24 so that the x3 values are equal
You need to see my second post ...
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#6
(Original post by Muttley79)
You need to see my second post ...
Yes thank you, I have just realised
0
4 weeks ago
#7
use the fact that the product of the roots is equal to -(-3)/24 = 8. You can deduce that a=0.5. Now you can deduce that 1/r + 1 + r = -(28)/24*0.5 . Multiply by r since r cannot be zero to get a quadratic in r. Solve this. Find the 2 possibilities for 3 roots, and check which one satisfies that the sums of the pairs is equal to -14/24.
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