To solve the equation x² - xq = 12, rearrange it to x² - xq - 12 = 0, then factor using algebraic identities (x² - 4 = (x-2)(x+2) and x³ + 8 = (x+2)(x² - 2x + 4)) to find one real root x = -2 and two complex roots x = (3 ± i√15)/2 using the quadratic formula.
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Maths Olympiad | Can you solve for x.Added:
In this question, one equation is given that is x² - x cq is equal to 12. And we have to find the value of x. So for this we can write this equation as a x² - xq is = 12.
Then I can write x² - xq bring this 12 left side. So this will become -2 then equal to 0 x² - xq. And I can write -12 as a -4 - 8 is equal to 0. Okay. Then after I can write x² this -4 together and - x cq and - 8 together we can write okay here x² - 4 can written as uh 2 ^ 2 - x cq 8 can be written as 2 ^ 3 is equal to 0.
Then after I can write here x² - 2 ^ 2 and I will take minus common from these two terms. So I can write here x cq + 2 ^ 3 is equal to 0. And here apply we will apply uh one formula for this actually I can write here a² minus b² is equal to what we can write a minus b and a + b same as I can write for this another formula okay that means a cq + b cq is equal 2 a + b a² - a b + b² we can write this formula.
So by using this two formula we can write here this as a x + 2 x - 2 minus we can write and this can be converted as a a + b that is x + 2 and uh this x² - a b that means 2 into x okay and plus b² that is b b is 2 so 2² is 4 we and write it okay is equal to zero after that if you will see this here x + 2 is and x + 2 in both term it is available so I can take this common so we can write x + 2 as a common so here will left only x - 2 and here minus and this term we left x² - 2 x + 4 is equal 0 then we can write as a x + 2 x - 2 - this will become - x² + 2x - 4 is equal to z okay then we can write here x + 2 here x you see this - x² so I will write - x² and 2x + x + 3x it will become -2 -4 - 6 it will become is equal to 0. So this means what we can write x + 2 is =0 and uh this we can write - x² + 3x - x =0 we can write. So from here x is = -2 first let's suppose this is x1 and from after uh we will solve this now. So we can after multiplying minus 1 both sides. So we can write this x² - 3x + 6 is = 0. And using this one formula that is x = - b + - under root b² - 4 a c is uh divided by 2 a. We can write here what for for this equation a is what here cofficient of x² that is one and b is the cofficient of x that is minus3 and c is here constant term that is 6.
So after putting the value of this we can write here x is equal to uh minus - -3 in b place of b we can write - 3 then after + minus under root b² that is - 3² - 4 into a a is 1 and c is 6 so we can write upon 2 into 1 okay so this will become 3 + - And this 3 - 3 square is a 9 and 4 into 6 - 24 which is an under root upon 2. Okay. Then we can write 3 + - 9 - 24 that is under root -5 upon 2. We can write this as a is equal to 3 + minus you know and we can write under root 15 as a -1 into under root that is 15 in this way. So this is actually what 3 + - under root minus 1 is called i. So we can write under root 15 i upon 2. So here also two value of x. So I can write x 2 is equal to what 3 + - under sorry + under<unk> 15 I that is I is iota here and x3 we can write here 3 minus under roo<unk> 15 iota upon 2 and we have already calculated x1 as a minus2 so you can write x1 as a minus2. So these are the required answer for this question.
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