The number of distinct real roots of
$x^{4} - 4 x^{3} + 12 x^{2} + x - 1 = 0$ is

0.0

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Solution

The correct option is C 2
We have $x^{4} - 4 x^{3} + 12 x^{2} + x - 1 = 0 \Rightarrow x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 + 6 x^{2} + 5 x - 2 = 0 \Rightarrow (x - 1)^{4} + 6 x^{2} + 5 x - 2 = 0 \Rightarrow (x - 1)^{4} = - 6 x^{2} - 5 x + 2$
To solve the above polynomial, it is equivalent to fine intersection points of the curves $y = (x - 1)^{4}$ and
$y = - 6 x^{2} - 5 x + 2 o r y = (x - 1)^{4}$ and $(x + \frac{5}{12})^{2} = - \frac{1}{6}$
The graph or above two curves as follows.
Clearly they have two points of intersection.
Hence the given polynomial has two real roots.

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x^{4} - 4 x^{3} + 12 x^{2} + x - 1 = 0

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f (x) = | 9 - x^{2} | - | x - a |

f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} .

s

f (x) = 0

| s | = t .

s

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