m = sqrt(x) m(m^2-1)=6 m^3-m-6=0 It's a cubic so we are expecting three m values By Rational Root Theorem, root candidates are +-1, +-2, +-3, +-6 By trial +2 is a root Because +2 is a root, (m-2) is a factor So m^3-m-6 = (m-2)q(m) So q(m) = (m^3-m-6)/(m-2) We can find q(m) using polynomial division By polynomial division, q(m)=m^2+2m+3 To find the roots of q(m), complete the square Completing the square: m^2+2m+1=-3+1=-2 (m+1)^2=-2 m+1 = +-sqrt(-2) m=-1+-sqrt(2)i We now have the three m values: {2,-1+sqrt(2)i,-1-sqrt(2)i} x = m^2 = {2^2, (-1+sqrt(2)i)^2, (-1-sqrt(2)i)^2} By FOIL, (-1+sqrt(2)i)^2 = (-1+sqrt(2)i)(-1+sqrt(2)i) = 1-sqrt(2)i-sqrt(2)i-2 = -1-2*sqrt(2)i Similarly, (-1-sqrt(2)i)^2 = -1+2*sqrt(2)i So x = {4,-1-2*sqrt(2)i,-1+2*sqrt(2)i} Notice I used no magic insight or "voila!" factoring. This was all just sheer calculation. If you are a student learning to solve these, go learn the Rational Root Theorem, go learn polynomial division, and go learn completing the square.
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Since what you did by factoring was essentially the same thing as polynomial long division why not just do polynomial long division directly?
For solving in less time! ❤
m = sqrt(x)
m(m^2-1)=6
m^3-m-6=0
It's a cubic so we are expecting three m values
By Rational Root Theorem, root candidates are +-1, +-2, +-3, +-6
By trial +2 is a root
Because +2 is a root, (m-2) is a factor
So m^3-m-6 = (m-2)q(m)
So q(m) = (m^3-m-6)/(m-2)
We can find q(m) using polynomial division
By polynomial division, q(m)=m^2+2m+3
To find the roots of q(m), complete the square
Completing the square: m^2+2m+1=-3+1=-2
(m+1)^2=-2
m+1 = +-sqrt(-2)
m=-1+-sqrt(2)i
We now have the three m values: {2,-1+sqrt(2)i,-1-sqrt(2)i}
x = m^2 = {2^2, (-1+sqrt(2)i)^2, (-1-sqrt(2)i)^2}
By FOIL, (-1+sqrt(2)i)^2 = (-1+sqrt(2)i)(-1+sqrt(2)i) = 1-sqrt(2)i-sqrt(2)i-2 = -1-2*sqrt(2)i
Similarly, (-1-sqrt(2)i)^2 = -1+2*sqrt(2)i
So x = {4,-1-2*sqrt(2)i,-1+2*sqrt(2)i}
Notice I used no magic insight or "voila!" factoring. This was all just sheer calculation.
If you are a student learning to solve these, go learn the Rational Root Theorem, go learn polynomial division, and go learn completing the square.
Very nice! ❤
Call √x = y; then the eqn. becomes ²³
y(y² -1) = 6.
y³ -y = 6.
y³ -y -6 = 0.
We need a cub term in the numeric, so 6 = 8 -2 = 2³ -2.
y³ -y -(2³ -2) = 0
y³ -2³- (y -2) = 0
(y -2)(y²+2y+4)- (y -2) = 0
(y -2)(y²+2y+4 -1) = 0
(y -2)(y²+2y+3) = 0
(y -2) = 0 and (y²+2y+3) = 0 give two solutions.
(y -2) = 0,
y=2 . . . . soln.1
y²+2y+3 = 0 is a quadratic eqn. where discriminant D = √[2² - 4 . 1 . 3] = √[-8] which doesn't give real roots.
∴ y=2=√x.
x=4.
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√4 (4-1)=6
2*3=6
6=6
:.x=4
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x(x^2_2x+1)=36=4.9 , where x=4, and x^2-2x+1=9
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√x * (x - 1) = 6 → let: m = √x
m.(m² - 1) = 6
m³ - m = 6
m³ - m = 8 - 2
m³ - 8 - m + 2 = 0
(m³ - 2³) - (m - 2) = 0 → recall: (a³ - b³) = (a - b).(a² + ab + b²)
(m - 2).(m² + 2m + 4) - (m - 2) = 0
(m - 2).[(m² + 2m + 4) - 1] = 0
(m - 2).(m² + 2m + 3) = 0
First case: (m - 2 ) = 0
m = 2 → recall: √x = 2
√x = 2
→ x = 4
Second case: (m² + 2m + 3) = 0
Δ = 2² - (4 * 3) = 4 - 12 = - 8 = 8i² = 4i² * 2
m = (- 2 ± 2i√2)/2
m = - 1 ± i√2 → recall: √x = 2
√x = - 1 ± i√2
x = (- 1 ± i√2)²
x = 1 ± 2i√2 + 2i²
x = 1 ± 2i√2 - 2
→ x = - 1 ± 2i√2
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Ровно за 3 секунды в уме решил что х равен 4. К чему эта демагогия??
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X=4
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let u=Vx , u^3+/-u^2-u-6=0 , (u-2)(u^2+2u+3)=0 , u=(-2+/-V(4-12))/2 , u=2 , Vx=2 , x=4 ,
1 -2 / u= -1+i*V2 , -1+i*V2 /,
2 -4
3 -6
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@@SALogics Thanks!
@@prollysine You're very welcome! ❤❤
@@SALogics Thanks!
x=4
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