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 used was $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find the value of $

Vedclass · Accepted Answer

$-3, -10$

Vedclass · Answer

(B) The incorrect equation used was $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find the value of $q$ using the product of roots formula: $q = (-2) 	imes (-15) = 30$. Now,substitute $q = 30$ and the correct coefficient of $x$ (which is $13$) into the original equation: $x^2 + 13x + 30 = 0$. To find the roots,factorize the quadratic equation: $x^2 + 10x + 3x + 30 = 0$. $x(x + 10) + 3(x + 10) = 0$. $(x + 3)(x + 10) = 0$. 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 $a^{2}=by+cz, b^{2}=cz+ax, c^{2}=ax+by,$ then the value of $\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$ is

Two positive distinct numbers $a$ and $b$ each differ from their reciprocal by $1$. The value of $a + b$ is

If one of the roots of the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is coincident,then the numerical value of $(a + b)$ is

Solve the given two equations and select the correct answer from the given options:
$I.$ $679 x^{2} - 168 x^{2} = 3066$
$II.$ $\sqrt{144} y^{3} - 9 y^{3} = 1536$

If the sum of the roots of the equation $x^2 + px + q = 0$ is three times their difference,then which one of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module