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

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(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$ $(

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$2$

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(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

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