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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

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

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

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

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

Join BYJU'S Learning Program