The equation whose roots are increased by $1$ than those of $x^{4}+1=0$, is:
- $x^{4}+1=0$
- $x^{4}-4x^{3}+6x^{2}-4x+2=0$
- $x^{4}+4x^{3}-6x^{2}-4x+2=0$
- $x^{4}-1=0$
Reveal answer
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B
Correct answer
Explanation
If the roots of x^4+1=0 are r, the new roots are r-1. Let y = x+1, so x = y-1. Substitute into the equation: (y-1)^4 + 1 = 0. (y^4 - 4y^3 + 6y^2 - 4y + 1) + 1 = 0. y^4 - 4y^3 + 6y^2 - 4y + 2 = 0.
AI explanation
To find an equation whose roots are each greater by 1 than those of the original equation, use the transformation by substituting (x - 1) for x in the given equation. The new equation becomes (x - 1)^4 + 1 = 0. Expand this using the binomial theorem to get x^4 - 4x^3 + 6x^2 - 4x + 1 + 1 = 0. Combine the constant terms to reach the final equation x^4 - 4x^3 + 6x^2 - 4x + 2 = 0.