The number of real roots of the polynomial eq | vedclass

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(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

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$2$

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(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

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