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anonymous
 one year ago
what is the answer..... (7x squared  2x+1) + (8x cubed + 2x squared +5x 4???
anonymous
 one year ago
what is the answer..... (7x squared  2x+1) + (8x cubed + 2x squared +5x 4???

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is a two step equation

Drigobri
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure about this numbers? If you try to calculating (7x²  2x + 1) = (ax²  bx +c) 1)Calculating Δ: Δ = b²  4.a.c Δ = 2²  4. 7. 1 Δ = 4  4 7. 1 Δ = 24 There are no real roots.

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0I think when you open the second bracket up and collect like terms, the determinant would change and real solutions will become a possibility

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@BAdhi I can't find any zeros for this cubic :/ I tried subbing in random points as well ask factoring... any ideas?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1I guess since you are asking for an answer it is an equation not jst a statement, So, \(f(x) = (7x^22x+1)+(8x^3+2x^2+5x4)\) \(f(x) = 8x^3+9x^2+3x3 \) when this is graphed a one we can see that it cuts the y = 0 axis at 0.389 so there is no way you gonna find that value by jst guessing values. for these kind of equations, there are several methods that can be followed. A famous method is the NewtonRaphson method which is an iterative method where you guess the first value which you think it is the root and with that value you can reach close to the actual root by iteration. I dont know why you say that this is two step question though : /

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@BAdhi oh okay, yeah I remember learning about that method. Thanks.
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