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Question

(A) Given equation: $(x+1)^{2} + x^{2} = 221$Expand the square: $(x^{2} + 2x + 1) + x^{2} = 221$Combine like terms: $2x^{2} + 2x + 1 = 221$Subtract $2

Vedclass · Accepted Answer

$10, -11$

Vedclass · Answer

(A) Given equation: $(x+1)^{2} + x^{2} = 221$Expand the square: $(x^{2} + 2x + 1) + x^{2} = 221$Combine like terms: $2x^{2} + 2x + 1 = 221$Subtract $221$ from both sides: $2x^{2} + 2x - 220 = 0$Divide the entire equation by $2$: $x^{2} + x - 110 = 0$Factorize the quadratic equation: $x^{2} + 11x - 10x - 110 = 0$Group the terms: $x(x + 11) - 10(x + 11) = 0$$(x - 10)(x + 11) = 0$Therefore,$x - 10 = 0$ or $x + 11 = 0$$x = 10$ or $x = -11$Thus,the solutions are $10, -11$.

Solve the following equation using the method of factorization: $(x+1)^{2} + x^{2} = 221$.

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