Why not first try factoring as: ( 2x² + x ± 1 ) ( 3x² - x ∓ 1 ) ? The linear coefficients in both are clear, and the problem form strongly suggests that both constants are units; then a little trial and error quickly sets the constants as respectively +1 and -1 giving ( 2x² + x + 1 ) ( 3x² - x - 1 ). Much simpler than the gyrations you went through. It's also important to note that quartic problems are highly likely to be biquadratic, amenable to this approach.
Why not first try factoring as:
( 2x² + x ± 1 ) ( 3x² - x ∓ 1 ) ?
The linear coefficients in both are clear, and the problem form strongly suggests that both constants are units; then a little trial and error quickly sets the constants as respectively +1 and -1 giving
( 2x² + x + 1 ) ( 3x² - x - 1 ).
Much simpler than the gyrations you went through.
It's also important to note that quartic problems are highly likely to be biquadratic, amenable to this approach.
{6x^4+6x^4 ➖ }+{x^2+x^2 ➖ }={12x^8+x^4}= 12x^12 3^4x^3^4 1^2^2^3^2^2 1^1^1x^3^1^2 x^3^2 (x ➖ 3x+2).
A very tricky Algebra Question: 6x⁴ + x³ = 2x + 1, x ϵ R; x =?
x ϵ R; No complex or imaginary value root
(6x⁴ + x³ - x²) + (x² - 2x) - 1 = (3x² - x)(2x² + x) + [(3x² - x) - (2x² + x)] - 1 = 0
(2x² + x + 1)(3x² - x - 1) = 0, 2x² + x + 1 = 0, x = (- 1 ± i√7)/4; Rejected, x ϵ R
3x² - x - 1 = 0, x = (1 ± √13)/6
Answer check:
x = (1 ± √13)/6, 3x² - x - 1 = 0: x² = (x + 1)/3
6x⁴ + x³ = x²(6x² + x) = (1/3)(x + 1)(2x + 2 + x) = (1/3)(x + 1)(3x + 2)
= (1/3)(3x² + 5x + 2) = (1/3)(x + 1 + 5x + 2) = (1/3)(6x + 3) = 2x + 1; Confirmed
Final answer:
x = (1 + √13)/6 or x = (1 - √13)/6
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You've wrong at the minute 14. you write (1/X= 2c/-b ......), while I think (2a/-b....) is corrects.
It's a special type of quadratic formula:ruclips.net/video/mDaXTf60ivo/видео.htmlsi=nWU_d7sJQDtYrkHZ
@@superacademy247 thank you so much. youre right.
I concur with that. He has solved for 1/X rather than X
Marne.
先にx≠0を示すべき。
Absolutely 💯