Maths Question
Discussion
I feel stupid asking this but I've decided to do a maths class and one of the questions in my homework has thrown me! It's not as easy once you finish school and have a full time job and travel. The question is as follows:
Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
Have you done quadratic equations yet? That's what this is.
As you say:
x^2 + (x + 3)^2 = 369
Expand the brackets
x^2 + x^2 +6x +9 = 369
so
2x^2 +6x = 360
Divide through by 2
x^2 +3x = 180
Rearranging
x^2 +3x -180 = 0
Now this is a standard quadratic equation (google it) with a=1, b=2, c=-180
So using the standard solution for a quadratic equation, x = -3 +/- sqrt (9-(4*-180))/2
Ignore the negative solution and you get x = (-3 + 27)/2 = 12
which matches your answer.
As you say:
x^2 + (x + 3)^2 = 369
Expand the brackets
x^2 + x^2 +6x +9 = 369
so
2x^2 +6x = 360
Divide through by 2
x^2 +3x = 180
Rearranging
x^2 +3x -180 = 0
Now this is a standard quadratic equation (google it) with a=1, b=2, c=-180
So using the standard solution for a quadratic equation, x = -3 +/- sqrt (9-(4*-180))/2
Ignore the negative solution and you get x = (-3 + 27)/2 = 12
which matches your answer.
TehRin said:
I feel stupid asking this but I've decided to do a maths class and one of the questions in my homework has thrown me! It's not as easy once you finish school and have a full time job and travel. The question is as follows:
Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
I'd start by expanding it out-Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
x^2 + x^2 + 6x + 9 = 369, then simplify it up a bit
x^2 + 3x - 180 = 0
Then, by inspection you could spot X is either 12 or 15 (15 - 12 = 3, 15 X -12 = -180) or, solve the quadratic equation using
X = (-b +/- sqrt (b^2/4ac))/2a
AmitG said:
Have you done quadratic equations yet? That's what this is.
As you say:
x^2 + (x + 3)^2 = 369
Expand the brackets
x^2 + x^2 +6x +9 = 369
so
2x^2 +6x = 360
Divide through by 2
x^2 +3x = 180
Rearranging
x^2 +3x -180 = 0
At this point you can just factorise the equation:As you say:
x^2 + (x + 3)^2 = 369
Expand the brackets
x^2 + x^2 +6x +9 = 369
so
2x^2 +6x = 360
Divide through by 2
x^2 +3x = 180
Rearranging
x^2 +3x -180 = 0
=> (x-12)(x+15) = 0
So either x+15 = 0 i.e x=-15 (but you can't have a negative age) or
x-12 = 0 i.e. x = 12 (as above)
QED
Zia = x
Raj = x+3
x^2 + (x+3)^2 = 369
Expand brackets (x+3)(x+3) = x^2 + 3x + 3x + 9
Collect like terms x^2 + x^2 + 3x + 3x + 9 = 369
2x^2 + 6x + 9 = 369
Simplify 2x^2 + 6x = 360
Factorise 2(x^2 + 3x) = 360
Divide both sides by 2 to simplify:
x^2 + 3x - 180 = 0
Solve by a guess or the quadratic foirmula
x = -15 or x = +12
the negative solution is NA, therefore x = 12
therefore..
Zia = 12
Raj = 12 + 3 = 15
Raj = x+3
x^2 + (x+3)^2 = 369
Expand brackets (x+3)(x+3) = x^2 + 3x + 3x + 9
Collect like terms x^2 + x^2 + 3x + 3x + 9 = 369
2x^2 + 6x + 9 = 369
Simplify 2x^2 + 6x = 360
Factorise 2(x^2 + 3x) = 360
Divide both sides by 2 to simplify:
x^2 + 3x - 180 = 0
Solve by a guess or the quadratic foirmula
x = -15 or x = +12
the negative solution is NA, therefore x = 12
therefore..
Zia = 12
Raj = 12 + 3 = 15
speedking31 said:
An option to calculate the answer without having to solve a quadratic.
Let x be their average age.
Raj = x + 1.5. Zia = x - 1.5.
Raj² = x² + 3x + 1.5²
Zia² = x² - 3x + 1.5²
Raj² + Zia² = 2x² + 4.5
369 = 2x² + 4.5
364.5 = 2x²
182.25 = x²
x = 13.5
Raj = 15, Zia = 12
That's brilliant. So simple yet it never would have thought of that in a million years. Like everyone else, went straight to quadratic. Let x be their average age.
Raj = x + 1.5. Zia = x - 1.5.
Raj² = x² + 3x + 1.5²
Zia² = x² - 3x + 1.5²
Raj² + Zia² = 2x² + 4.5
369 = 2x² + 4.5
364.5 = 2x²
182.25 = x²
x = 13.5
Raj = 15, Zia = 12
Love these threads, very nostalgic.
Yep, given the low total of 369 there aren't many options so I'm with Strudul and an educated guess is the way to go here especially for only 2 points (keep trying to teach my kids time management for exams).
However, Speedking's answer is fantastic. Never seen that approach before and will add it to the locker. Great stuff.
Yep, given the low total of 369 there aren't many options so I'm with Strudul and an educated guess is the way to go here especially for only 2 points (keep trying to teach my kids time management for exams).
However, Speedking's answer is fantastic. Never seen that approach before and will add it to the locker. Great stuff.
Shakermaker said:
Pythagorus taught us also that
A squared + B squared = C squared
But I don't know how to use that to your advantage in this situation given that the answers have already been given...
Unless I'm missing something, maybe a parrot, I think that only applies to right angled triangles where C is the hypotenuse. So not sure it's key to this question. A squared + B squared = C squared
But I don't know how to use that to your advantage in this situation given that the answers have already been given...
TehRin said:
I feel stupid asking this but I've decided to do a maths class and one of the questions in my homework has thrown me! It's not as easy once you finish school and have a full time job and travel. The question is as follows:
Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
Just put bloody unchecked immigration Raj is three years older than Zia.
The sum of the squares of their ages is 369.
How old are they?
It's only a 2 mark question and the answer an be obtained through an educated guess and/or trial and error.
Raj = 15
Zia = 12
The question is basically:
x^2 + (x + 3)^2 = 369
What formula/method can I use to work this out.
TwigtheWonderkid said:
Shakermaker said:
Pythagorus taught us also that
A squared + B squared = C squared
But I don't know how to use that to your advantage in this situation given that the answers have already been given...
Unless I'm missing something, maybe a parrot, I think that only applies to right angled triangles where C is the hypotenuse. So not sure it's key to this question. A squared + B squared = C squared
But I don't know how to use that to your advantage in this situation given that the answers have already been given...
Some elegant stuff above.
But for just two marks, in an exam, speed is of the essence.
I'd have started with 10 squared, and gone up from there, and within a few seconds, you'll find the two numbers separated by 3 that give you 369.
The small number of marks is a clue that there's a short-cut. If you can't see the short-cut straight away, leave it to the end.
But for just two marks, in an exam, speed is of the essence.
I'd have started with 10 squared, and gone up from there, and within a few seconds, you'll find the two numbers separated by 3 that give you 369.
The small number of marks is a clue that there's a short-cut. If you can't see the short-cut straight away, leave it to the end.
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