The solution set of a quadratic equation x^{2 | vedclass

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

Question

(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.We look for two numbers whose product is

Vedclass · Accepted Answer

$\{-7, 2\}$

Vedclass · Answer

(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.We look for two numbers whose product is $-14$ and whose sum is $5$.These numbers are $7$ and $-2$,since $7 	imes (-2) = -14$ and $7 + (-2) = 5$.Thus,we can write the equation as: $x^{2} + 7x - 2x - 14 = 0$.Grouping the terms,we get: $x(x + 7) - 2(x + 7) = 0$.Factoring out $(x + 7)$,we have: $(x + 7)(x - 2) = 0$.Setting each factor to zero,we get $x + 7 = 0$ or $x - 2 = 0$.Therefore,$x = -7$ or $x = 2$.The solution set is $\{-7, 2\}$.

The solution set of a quadratic equation $x^{2}+5x-14=0$ is $\ldots \ldots \ldots \ldots .$

Similar Questions

Verify whether the given value of $x = 2$ is a solution of the quadratic equation $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$ or not.

Solve the following equation using the method of factorization: $\frac{x^{2}-1}{x^{2}+1}=\frac{7}{9}$

We get $2$ litres of kerosene less in Rs. $360$ if the price of kerosene increases by Rs. $2$ per litre. Find the original price of kerosene per litre.

Solve the following equation by using the quadratic formula,if the equation has a solution in $R$: $x^{2}-3x-4=0$.

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^{2} - 6x + 3 = 0$; $x = \frac{1}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module