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Answer :
So there are 3 siblings: the first twin, the second twin, and their older sister.
Since the sister is older by 11 years, you can represent their relationship with the following equation;
age of twin = age of older sister - 11
t = s - 11
Representing this next part may be a little hard to show, but i'll do my best :)
Adding the ages together can be represented like this;
total age = t + t + s
If the product of their ages ( t * t * s ) is 1181 more than the sum, you can show it like this:
[tex]t * t * s = (t + t + s) + 1181[/tex]
Now you substitute in (s - 11) for t because s = t - 11.
[tex]t * t * s = (t + t + s) + 1181\\
(s-11) * (s-11) * s = ((s-11) + (s-11) + s) + 1181\\
(s^{2}-22x+121)*s=s+s-11-11+1181\\
s^{3}-22s^{2}+121s-2s=1159\\
s^{3}-22s^{2}+118s-1159=0[/tex]
Factor:
Since it's a third-degree equation, I just entered it into a polynomial solver in my calculator. There was only one factor listed, x = 19. Using (s - 19) = 0 as a factor, the equation becomes the following:
[tex]s^{3}-22s^{2}+118s-1159=0\\
(s-19)(s^{2}-3s+61)=0[/tex]
Using the quadratic formula, you get:
[tex]s= \frac{-b(+-)\sqrt{b^{2}-4ac}}{2a}\\
s= \frac{-(-3)(+-)\sqrt{(-3)^{2}-4(1)(61)}}{2*1}\\
s= \frac{3(+-)\sqrt{(9)-244}}{2}\\
s= \frac{3(+-)\sqrt{-235}}{2}[/tex]
Since you have a square root of a negative number, the answer will be imaginary. You cannot have an age that is imaginary, so the age of the older sister (s) is 19 years.
Because t = s - 11, that means that t = (19) - 11 = 8.
t is the age of the twins, so the twins are 8 years old.
Since the sister is older by 11 years, you can represent their relationship with the following equation;
age of twin = age of older sister - 11
t = s - 11
Representing this next part may be a little hard to show, but i'll do my best :)
Adding the ages together can be represented like this;
total age = t + t + s
If the product of their ages ( t * t * s ) is 1181 more than the sum, you can show it like this:
[tex]t * t * s = (t + t + s) + 1181[/tex]
Now you substitute in (s - 11) for t because s = t - 11.
[tex]t * t * s = (t + t + s) + 1181\\
(s-11) * (s-11) * s = ((s-11) + (s-11) + s) + 1181\\
(s^{2}-22x+121)*s=s+s-11-11+1181\\
s^{3}-22s^{2}+121s-2s=1159\\
s^{3}-22s^{2}+118s-1159=0[/tex]
Factor:
Since it's a third-degree equation, I just entered it into a polynomial solver in my calculator. There was only one factor listed, x = 19. Using (s - 19) = 0 as a factor, the equation becomes the following:
[tex]s^{3}-22s^{2}+118s-1159=0\\
(s-19)(s^{2}-3s+61)=0[/tex]
Using the quadratic formula, you get:
[tex]s= \frac{-b(+-)\sqrt{b^{2}-4ac}}{2a}\\
s= \frac{-(-3)(+-)\sqrt{(-3)^{2}-4(1)(61)}}{2*1}\\
s= \frac{3(+-)\sqrt{(9)-244}}{2}\\
s= \frac{3(+-)\sqrt{-235}}{2}[/tex]
Since you have a square root of a negative number, the answer will be imaginary. You cannot have an age that is imaginary, so the age of the older sister (s) is 19 years.
Because t = s - 11, that means that t = (19) - 11 = 8.
t is the age of the twins, so the twins are 8 years old.
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