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College interview question that confused most students
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- Опубликовано: 25 авг 2024
- What do you think about this question? If you're reading this ❤️.
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It is important to recognise that this is a cubic equation and therefore there MUST be three solutions. If you don't remember the difference of two cubes formula, it is easy enough to spot that x=3 is a solution, so (x-3) must be a factor - then divide x^3 +0x^2+0x-27 by x-3.
There's no need to make things simple if you can make them complicated.
actually saw this in a interview for a math heavy computer science interview. i asked the interviewer if we were interested in all roots or just real roots and he he said very good then moved to the next question. one of my first interviews and it kinda freaked me out. but i guess they just wanted to see if i knew the answer wasn't just 3 and move on.
x³ = 27 = 3³*e^(i*2*π*k), k integer
x = 3*e^(i*2*π*k/3)
x = 3, 3*(-1 ± i*√3)/2
These are 3 times the cube roots of 1.
If the domain is not defined, there is no reason to go beyond real solutions.
Hello. Here is an alternate solution. The answer can we worked out without the complicated mathematics. We know that the simple answer is 3 but how to find the complex roots. Just draw the vector with a modulus of 3 in the complex plane. (Remember when cubing the vector the angle is multiplied by 3.) The first vector is aligned with Zero degrees 3 x 0 = 0. The second vector is aligned with 120 degrees as 3 x 120 degrees is 360 degrees or zero degrees.. The last answer is a bit trickier. The vector is aligned with 240 degrees as 3 x 240 = 720 or zero. The answers are the complex coordinates of the vectors. Ans1. 3 + 0i, Ans 2. -1.5 + {(3 x 3^0.5)/2}i, Ans3. -1.5 - {(3 x 3^0.5)/2}i I find it easier to visualise the problem when drawing the problem in the complex plane. This method helps significantly when higher exponents are used ...say 12
x³ = 27 = 3³*e^(i*2*π*k), k integer
x = 3*e^(i*2*π*k/3)
x = 3, 3*(-1 ± i*√3)/2
These are 3 times the cube roots of 1.
Why make it so cpomplicated? Inthe very first step, if you know that x`3 (read x-cubed) minus 3`3 = 0, then x must equal 3, and 27 - 27 = 0. YOur question did not say, "for all x," so the answer 3 is sufficient.
when you need both real and imaginary roots of the solutions ❤
Your basic premise for objection is correct, but please permit me to phrase it differently. It is nothing to do with complications or the lack thereof, but in the accurate use of language. The complications follow in the determination of the complex roots, but they need not be sought due to the phrasing of the question. Mathematics is a discipline of accuracy, and deserves such in the stipulation of required response, somewhat lacking here.
I agree with you first sentiments...and I had always felt this way about life...why make things more complicated then they are. However, I almost died in a hospital 5 years ago because the questions posed to the doctor about my condition were clearly answered by the test results. I was recovering from a liver abscess, and told to go home. My tests all showed me gaining my health. My wife looked at me, and told the doctor she wanted another CAT scan....she could feel, that I was not as healthy as the test results showed. The radiologist that treated me upon further testing could not believe that the doctor had not also done a clinical evaluation of my condition, and not just the obvious. I believe math is at the center of this type of problem solving.
it also doesn't say find a x. the question needs clarification. had this come up in a job interview. asked if they wanted just a root, just real roots, or all roots. the guy just said good and moved to the next question
Interview questions are designed to make you think. Getting the obvious answer means you don’t get the job. I for one wouldn’t employ you. Next!
And that’s why I got a first class degree from one of the best universities in the UK and others didn’t even get a place on the course.
It really does matter HOW you think: getting an answer is not good enough, it just shows you really don’t understand the question or think about what the less obvious answers could be. Interview questions are designed to find out HOW you think. It doesn’t even matter if you get the wrong answer, what matters is that you think about the question in the right way. You obviously didn’t. I did!
x³ = 27 = 3³*e^(i*2*π*k), k integer
x = 3*e^(i*2*π*k/3)
x = 3, 3*(-1 ± i*√3)/2
These are 3 times the cube roots of 1.
3, 3∠120°, 3∠-120°
You’re right. Using the polar coordinates, x = 3 e^i thêta, here we can say that the 3 solutions are on a circle with radius = 3 and thêta = 0, 120 dg and -120 dg. Nice representation of the equation results !
Why you did it in a long method????? It was just easy!
This right here is why people tend to despise math.
I would have written 3 root 2 over 2 i, always keep i at the end of the complex number.
Anyway, my approach would be to find 3, plot on complex plane and find the other two roots on the radius 3 circle by dividing 360° by 3, producing two roots with negative real components and +/- imaginary components. You don’t need to remember equations, just think about complex numbers and use Euler’s formula.
But what you'll write isn't correct.
Good video explaining the complete thinking.
Thanks for sharing!
Nice Explanation Sir❤
i ❤ Mathematics
X3-27=0
X3=27
X3 = 33
X= 3
Learn to use the cube symbol, as in x³ or x^3. Otherwise, you're wrong. Also, there are 2 other unique solutions you missed. You'd only get 1/4 the score if I had graded your answer.
What do you mean
(x-3)(x^2+3x+9)=0
x-3=0 x=3
x^2+3x+9=0
x=[-3+-rq(9-36)]/2
x=[-3+-rq(-27)]/2
x=[-3+-3i•rq3]/2
X=3
You missed the other 2 unique solutions.
(1) x³ - 27 = 0 | -27
(2) x³ = 27 | apply log base 3
(3) 3 × log(x) = 3 | ÷ 3
(4) log(x) = 1
(5) x = 3
Please remember: 0nly if the x in log(x) is the base of this log then log(x) equals one.
Verification:
(1) x³ - 27 = 0 | x -> 3; -27
(2) 3³ = 27 = 3 × 3 × 3
Appears to be correct.
For the magnitude, just take the cube root of 27 to get 3. Then remember to consider the 3 cube roots of 1, which are 1, and (-1 ± i*√3)/2. If you used polar coordinates on the complex plane, then it's easy to see they're positioned at 0⁰ and ±120⁰ at distance 3 from the origin.
Your first line should have "| +27".