If one of the roots of x^{2} + bx - 12 = 0 is | vedclass

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

Question

(B) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.Here,the equation is $x^{2} + bx - 1

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(B) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.Here,the equation is $x^{2} + bx - 12 = 0$,where $a = 1$ and $c = -12$.Let the roots be $\alpha$ and $\beta$. We are given $\alpha = 2$.The product of the roots $\alpha \cdot \beta = \frac{c}{a} = \frac{-12}{1} = -12$.Substituting the value of $\alpha$:$2 \cdot \beta = -12$.$\beta = \frac{-12}{2} = -6$.Thus,the other root is $-6$.

If one of the roots of $x^{2} + bx - 12 = 0$ is $2$,the other root is ............

Similar Questions

State whether the following statements are true or false. Justify your answers.
$(i)$ Every quadratic equation has exactly one root.
$(ii)$ Every quadratic equation has at least one real root.

The roots of the quadratic equation $\ldots \ldots \ldots \ldots$ are $-4$ and $5$.

Examine whether the following equation is quadratic or not: $7x = 2x^2$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-3}+\frac{x-3}{x+1}=\frac{5}{2}$

The product of the digits of a two-digit number is $14$. The number obtained by interchanging the digits is $45$ more than the original number. Find the original number.

Vedclass Test Series

Exam Paper Generator

Online Exam Module