Solve the following equation using the method | vedclass

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

Question

(B) Given equation: $x^{2} + (x + 5)^{2} = 125$Expand the term $(x + 5)^{2}$ using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$:$x^{2} + (x^{2} +

Vedclass · Accepted Answer

$5, -10$

Vedclass · Answer

(B) Given equation: $x^{2} + (x + 5)^{2} = 125$Expand the term $(x + 5)^{2}$ using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$:$x^{2} + (x^{2} + 10x + 25) = 125$Combine like terms:$2x^{2} + 10x + 25 = 125$Subtract $125$ from both sides to set the equation to zero:$2x^{2} + 10x - 100 = 0$Divide the entire equation by $2$ to simplify:$x^{2} + 5x - 50 = 0$Factorize the quadratic equation by finding two numbers that multiply to $-50$ and add to $5$:These numbers are $10$ and $-5$:$x^{2} + 10x - 5x - 50 = 0$$x(x + 10) - 5(x + 10) = 0$$(x - 5)(x + 10) = 0$Setting each factor to zero:$x - 5 = 0 \implies x = 5$$x + 10 = 0 \implies x = -10$Thus,the solutions are $5, -10$.

Solve the following equation using the method of factorization: $x^{2} + (x + 5)^{2} = 125$

Similar Questions

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $x^{2}-5x-1=0$

Solve the following equation using the method of factorization and write its solution set: $4x^{2} + 4x = 15$.

If the roots of the following quadratic equation exist,find them by the method of completing the square: $x + \frac{2}{x} - 8 = 0$.

If $\ldots \ldots \ldots$ then real roots of a quadratic equation do not exist.

Vedclass Test Series

Exam Paper Generator

Online Exam Module