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 roots of the quadratic equation by using the quadratic formula:
$-3x^{2} + 5x + 12 = 0$

Obtain the roots of the following equation using the method of 'completing the square': $3y^{2} + 7y - 20 = 0$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $x(4 - kx) = 3 - 2x$.

State whether the quadratic equation $2 x^{2}-6 x+\frac{9}{2}=0$ has two distinct real roots. Justify your answer.

If $3$ is a root of the quadratic equation $x^{2}-kx+6=0$,then $k = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module