Ok so I have this math assigment, but I'm stuck on one question, or maybe 2, but here one to start things off. Find the quadratic equation whose roots are the cubes of the roots of 3x^2-2x+9=0 Any help?
Dont have time as its 1am to do the maths but find the factorising of the equation. Find the roots. Then cube the roots. And use the formula you should know about constructing an equation given the roots ( will do maths if I wake up erly enough and you havent got an answer)
Yeah, I asked him if I could do that, but he's says it's cheap, lol. Probably should have mentioned that. He wants us to use some method where X^2-(Sum)X+(Product)= 0 Not sure if you know what I'm talking about. I don't really get it, cause I wasnt there for that lesson.
"X^2-(Sum)X+(Product)= 0" I'm guessing he just means the simple idea behind factoring. (x+y)(x+z)=0 y+z= "Sum" y*z= "Product" This type of thinking is good for this problem because the roots are imaginary. well you have 3x^2-2x+9=0 you can go about this two ways (3x-?)(x-?*)=0 3(x^2-2/3x+3)= 3(x-?)(x-?*)=0 The second one is easier in this case because you should recognize right away that there cannot be two negative numbers that will add up to -2/3 but multiply to 3, and therefore there are no real roots and only complex roots. There are two complex roots, because this is a 2nd degree polynomial with 0 real roots. using the quadratic equation, you get -(-2)+- sqrt((-2)^2-4(3)(9)) / 2(3) 2 +- sqrt (-104) / 6 (2 +- sqrt (-4) *sqrt (26))/6 (2 +- 2i * sqrt(26))/6 (1+ i sqrt(26))/3 and (1- i sqrt(26))/3 are the two roots complex roots Taking these roots, you just cube them. easier to do numerator and denominator separately. 3 cube = 27 1+i sqrt(26) * 1 + i sqrt(26) = 1 + 2isqrt^26) + -26 = 2isqrt(26) - 25 1-i sqrt(26) * 1 - i sqrt(26) = 1 - 2isqrt^26) + -26 = -2isqrt(26) - 25 ( 2isqrt(26) - 25 ) * (1 + i sqrt(26) ) = 2isqrt(26) + -2(26) -25 - 25isqrt(26) = -77 - 23isqrt 26 ( -2isqrt(26) - 25 ) * (1 - i sqrt(26) ) = -2isqrt(26) + -2(26) -25 + 25isqrt(26) = -77 + 23isqrt 26 the roots of the new polynomial are ( -77 + 23i sqrt(26) ) / 27 ( -77 - 23i sqrt(26) ) / 27 (Notice that the complex roots of the original euqation occured in conjugate pairs, so cubing just one would have saved time and been good enough to find the other) to find the equation of this new polynomial multiply (easier to get rid of the denominator of the complex root portion) ( x - ( -2079 + 621i sqrt(26)))* (x - ( -2079 - 621i sqrt(26))) x^2 + 2079x + 621i sqrt(26)x + 2079x + 4322241 + 1291059isqrt(26) - 621i sqrt(26)x - 1291059isqrt(26) + 10026666 x^2 + 4158x + 14348907 Enjoy