The value of a for which the equations x^2 - | vedclass

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

Question

(D) Let the common root be $\alpha$. Since $\alpha$ satisfies both equations,we have:$\alpha^2 - 3\alpha + a = 0$ $(i)$$\alpha^2 + a\alpha - 3 = 0$ $(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let the common root be $\alpha$. Since $\alpha$ satisfies both equations,we have:$\alpha^2 - 3\alpha + a = 0$ $(i)$$\alpha^2 + a\alpha - 3 = 0$ $(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(\alpha^2 - 3\alpha + a) - (\alpha^2 + a\alpha - 3) = 0$$-3\alpha - a\alpha + a + 3 = 0$$-\alpha(a + 3) + (a + 3) = 0$$(a + 3)(1 - \alpha) = 0$This implies either $a = -3$ or $\alpha = 1$.If $a = -3$,the equations become $x^2 - 3x - 3 = 0$ and $x^2 - 3x - 3 = 0$,which are identical. However,for distinct quadratic equations to have a common root,we usually look for the value of $a$ that satisfies the condition for a specific root.Substituting $\alpha = 1$ into equation $(i)$:$1^2 - 3(1) + a = 0$$1 - 3 + a = 0$$-2 + a = 0$$a = 2$.

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

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $36 x^{4}+369 x^{2}+900=0$
$II.$ $144 y^{4}+337 y^{2}+144=0$

If $x = 2 + 2^{2/3} + 2^{1/3}$,then $x^3 - 6x^2 + 6x = $

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{4} - 227 = 398$
$II.$ $y^{2} + 321 = 346$

If the product of the roots of the equation $2x^2 + 6x + \alpha^2 + 1 = 0$ is $-\alpha$, then the value of $\alpha$ will be

If $x^{4} + \frac{1}{x^{4}} = 119$,then the value of $x^{3} - \frac{1}{x^{3}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module