The coefficient of x in the equation x^2 + px | vedclass

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

Question

(B) The incorrect equation is $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find $q$ using the produc

Vedclass · Accepted Answer

$-3, -10$

Vedclass · Answer

(B) The incorrect equation is $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find $q$ using the product of roots: $q = (-2) 	imes (-15) = 30$. Now,substitute $q = 30$ into the original equation where the coefficient of $x$ was $13$: $x^2 + 13x + 30 = 0$. To find the roots of the original equation,factorize the quadratic: $x^2 + 10x + 3x + 30 = 0$ $x(x + 10) + 3(x + 10) = 0$ $(x + 3)(x + 10) = 0$. Thus,the roots are $x = -3$ and $x = -10$.

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

Similar Questions

If the charge on a capacitor is increased by $2 \, C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor (in coulomb) is

$H_{3}PO_{3}$ can be represented by structures $1$ and $2$ shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing $H_{3}PO_{3}$? If not,give reasons for the same.

During which phase of mitosis do the nuclear envelope and nucleolus completely disappear?

VedClass-Data Entry

Two point charges $-q$ and $+q$ are located at points $(0, 0, -a)$ and $(0, 0, a)$ respectively. The potential at a point $(0, 0, z)$ where $z > a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module