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

While selling a pen for Rs. $24$, the loss percentage is equal to its cost price in rupees. Find the cost price of the pen.

What is the value of $k$ if one of the roots of the quadratic equation $x^{2} + 6x + k = 0$ is $-4$?

Which of the following is a quadratic equation?

If both the roots of a quadratic equation $ax^2 + bx + c = 0$ $(a \neq 0)$ are equal,then ..... .

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module