If alpha, beta, gamma are the roots of the eq | vedclass

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

Question

(A) Let $y = 2\alpha + 1$,which implies $\alpha = \frac{y - 1}{2}$.Since $\alpha$ is a root of $x^3 + 2x - 5 = 0$,we substitute $\alpha$ into the equa

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(A) Let $y = 2\alpha + 1$,which implies $\alpha = \frac{y - 1}{2}$.Since $\alpha$ is a root of $x^3 + 2x - 5 = 0$,we substitute $\alpha$ into the equation:$\left(\frac{y - 1}{2}\right)^3 + 2\left(\frac{y - 1}{2}\right) - 5 = 0$$\frac{y^3 - 3y^2 + 3y - 1}{8} + (y - 1) - 5 = 0$Multiply the entire equation by $8$:$y^3 - 3y^2 + 3y - 1 + 8(y - 6) = 0$$y^3 - 3y^2 + 3y - 1 + 8y - 48 = 0$$y^3 - 3y^2 + 11y - 49 = 0$Comparing this with $x^3 + bx^2 + cx + d = 0$,we get $b = -3, c = 11, d = -49$.Thus,$|b + c + d| = |-3 + 11 - 49| = |-41| = 41$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 2x - 5 = 0$ and the equation $x^3 + bx^2 + cx + d = 0$ has roots $2\alpha + 1, 2\beta + 1, 2\gamma + 1$,then the value of $|b + c + d|$ is (where $b, c, d$ are constants):

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $(x+y)^{2} = 3136$
$II.$ $y+2513 = 2569$

The value of $a$ for which the equations $x^2 - 3x + a = 0$ and $x^2 + ax - 3 = 0$ have a common root is

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{1225} x + \sqrt{4900} = 0$
$II.$ $(81)^{1/4} y + (343)^{1/3} = 0$

The two roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ are in the ratio $3 : 2$. The roots are:

Solve the given two equations and choose the correct answer from the given options.
$I.$ $x^{2}+12x+36=0$
$II.$ $y^{2}+15y+56=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module