One of the real roots of the equation x3-6 x2+6 x-2=0 is

Question

One of the real roots of the equation x3-6 x2+6 x-2=0 is (A) -1 (B) 2 (C) (2(1/3)/2(1)3-1) (D) (2(1/3)/2(1)3+1)

Tardigrade Admin · Accepted Answer

We have, x3-6 x2+6 x-2=0 Let f(x)=x3-6 x2+6 x-2 f(4)=64-96+24-2=-10<0 f(5)=125-150+30-2=3>0 ∴ Root lie between 4 and 5 . Hence, (21 / 3/21 / 3-1) is lie between 4 and 5 . ∴ (2(1/3)/2(1)3-1) is one real root of x3-6 x2+6 x-2=0

Q. One of the real roots of the equation $x^{3} - 6 x^{2} + 6 x - 2 = 0$ is

$- 1$

$2$

$\frac{2 ^{\frac{1}{3}}}{2 ^{\frac{1}{3}} - 1}$

$\frac{2 ^{\frac{1}{3}}}{2 ^{\frac{1}{3}} + 1}$

Q. One of the real roots of the equation x3−6x2+6x−2=0 is

−1

2

231​−1231​​

231​+1231​​

We have, x3−6x2+6x−2=0 Let f(x)=x3−6x2+6x−2 f(4)=64−96+24−2=−10<0 f(5)=125−150+30−2=3>0 ∴ Root lie between 4 and 5 . Hence, 21/3−121/3​ is lie between 4 and 5 . ∴231​−1231​​ is one real root of x3−6x2+6x−2=0

Q. One of the real roots of the equation $x^{3} - 6 x^{2} + 6 x - 2 = 0$ is

$- 1$

$2$

$\frac{2 ^{\frac{1}{3}}}{2 ^{\frac{1}{3}} - 1}$

$\frac{2 ^{\frac{1}{3}}}{2 ^{\frac{1}{3}} + 1}$