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Algebra Level 4

Find the product of real roots of x 4 2 x 3 + x = 380 x^4-2x^3+x=380


The answer is -20.

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2 solutions

Chew-Seong Cheong
Apr 29, 2015

Yes. 挺好 means very good.

x 4 2 x 3 + x 380 = ( x 5 ) ( x 3 + 3 x 2 + 15 x + 76 ) = ( x 5 ) ( x + 4 ) ( x 2 x + 19 ) = 0 \begin{aligned} x^4-2x^3+x-380 & = (x-5)(x^3+3x^2+15x+76) \\ & = (x-5)(x+4)(x^2 -x+19) = 0 \end{aligned}

Therefore, the real roots are 4 -4 and 5 5 , and their product is 20 \boxed{-20} .

Did you use trial and error to find 5 first and then 4 ? Or is there any specific method to find them ?

Vighnesh Raut - 6 years, 1 month ago

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how do we use error to find 5 without a calculator?

Lorraine Xie - 6 years, 1 month ago

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He said Trial and Error \text{Trial and Error} . Which means bashing!

Pranjal Jain - 6 years, 1 month ago

No other method other than try and error. Check out useful tips from Niranjan Kanderia here. For those who know me slightly better would know that I use Excel spreadsheet to plot the graph.

Chew-Seong Cheong - 6 years, 1 month ago

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You can use Rational Zero theorem or Synthetic Division in factoring such equation...

Dale Angelo Cortez - 6 years ago

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@Dale Angelo Cortez Why don't you show it. I would like to know how and others may too.

Chew-Seong Cheong - 6 years ago

I do not know how to post here lol ,,My head aches due to the brackets,braces etc that are being used on posting.

Dale Angelo Cortez - 6 years ago

The product of all roots is 380 = 2 2 5 19 380=2 \cdot 2 \cdot 5 \cdot 19 . So if 19 19 is the product of real roots, and roots are integers since coefficient of fourth power is 1 1 , x = ± 1 , m u s t b e a r o o t . f ( ± 1 ) 0. S o t r y 2 , 2 , 4 , 4 , 5 , 5 , f ( ± 2 ) 0. f ( 4 ) 0. f ( 4 ) = 0 f ( 5 ) = 0. p r o d u c t = 20 x=\pm 1, ~must~~ be~ a ~root.~~~~ f(\pm 1)~\ne 0. ~~~So ~try~ 2,-2,4,-4,5,-5,~~~~f(\pm 2)~\ne 0.\\f(4)~\ne 0.~~~f(-4)~= 0 ~~~f(5)=0.~~~~~~product=\color{#D61F06}{\boxed{-20} }

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