Solve the following equation using the method | vedclass

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

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: $(x+1)^{2} + x^{2} = 221$.

Similar Questions

Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $x^{2} + x + \frac{1}{4} = 0$.

The sum of a natural number and its reciprocal is $\frac{5}{2}$. Then the number is $\ldots \ldots .$

Solve the following equation using the method of factorization: $6(2x+1)^2 = (2x+1) + 5$

If the roots of the following quadratic equation exist,find them by the method of completing the square: $(2x + 1) - \frac{4}{(2x + 1)} - 3 = 0$.

If the discriminant $D = 0$,then the roots of the quadratic equation $ax^2 + bx + c = 0$ are ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module