The number of real roots of the polynomial eq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $x^4-x^2+2x-1=0$Rewrite the equation as: $x^4-(x^2-2x+1)=0$This simplifies to: $x^4-(x-1)^2=0$Using the difference of squares form

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $x^4-x^2+2x-1=0$Rewrite the equation as: $x^4-(x^2-2x+1)=0$This simplifies to: $x^4-(x-1)^2=0$Using the difference of squares formula $a^2-b^2=(a-b)(a+b)$,we get:$(x^2-(x-1))(x^2+(x-1))=0$$(x^2-x+1)(x^2+x-1)=0$Case $1$: $x^2-x+1=0$. The discriminant $D = (-1)^2 - 4(1)(1) = 1-4 = -3$. Since $D Case $2$: $x^2+x-1=0$. The discriminant $D = (1)^2 - 4(1)(-1) = 1+4 = 5$. Since $D > 0$,there are two distinct real roots.Therefore,the total number of real roots is $2$.

The number of real roots of the polynomial equation $x^4-x^2+2x-1=0$ is

Similar Questions

Solve the equation $2x^{2}+x+1=0$.

If $(x-2)$ is a common factor of the expressions $x^2+ax+b$ and $x^2+cx+d$,then $\frac{b-d}{c-a}$ is equal to

Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then,for any choice of positive integers $a$ and $b$,$t^3$ is never equal to:

The expression $y = ax^2 + bx + c$ always has the same sign as $c$ if

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module