Two students,while solving a quadratic equati | vedclass

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

Question

(D) Let the correct quadratic equation be $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.The first student copied the constant term incorrectly,b

Vedclass · Accepted Answer

$6, -1$

Vedclass · Answer

(D) Let the correct quadratic equation be $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.The first student copied the constant term incorrectly,but the coefficient of $x^2$ $(a=1)$ and the coefficient of $x$ $(b)$ were correct. The sum of roots is $-b/a$. Since the roots were $3$ and $2$,the sum is $3 + 2 = 5$. Thus,$-b/1 = 5$,which means $b = -5$.The second student copied the constant term $(c = -6)$ and the coefficient of $x^2$ $(a = 1)$ correctly. The product of roots is $c/a = -6/1 = -6$.So,we have the equation $x^2 - 5x - 6 = 0$.Factoring the equation: $x^2 - 6x + x - 6 = 0$.$x(x - 6) + 1(x - 6) = 0$.$(x - 6)(x + 1) = 0$.Therefore,the correct roots are $x = 6$ and $x = -1$.

Two students,while solving a quadratic equation in $x$,one copied the constant term incorrectly and got the roots $3$ and $2$. The other copied the constant term and coefficient of $x^2$ correctly as $-6$ and $1$ respectively. The correct roots are:

Similar Questions

For what values of $k$ will the equation $x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0$ have equal roots?

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$
$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$

If the equations $x^2 + px + 2q = 0$ and $x^2 + qx + 2p = 0$ $(p \ne q)$ have a common root,then the value of $p + q$ is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,then the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is

Solve the given two equations and provide the correct answer from the given options.
$I.$ $\frac{x}{x+7} + \frac{x+7}{x} = 12$
$II.$ $\frac{y}{y+8} + \frac{y+8}{y} = 16$

Vedclass Test Series

Exam Paper Generator

Online Exam Module