If {x^2} + ax + 10 = 0 and {x^2} + bx - 10 = | vedclass

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(D) Let $\alpha$ be a common root of the two equations.Then,${\alpha ^2} + a\alpha + 10 = 0$ $(i)$ and ${\alpha ^2} + b\alpha - 10 = 0$ $(ii)$.Subtrac

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) Let $\alpha$ be a common root of the two equations.Then,${\alpha ^2} + a\alpha + 10 = 0$ $(i)$ and ${\alpha ^2} + b\alpha - 10 = 0$ $(ii)$.Subtracting $(ii)$ from $(i)$,we get: $(a - b)\alpha + 20 = 0$,which implies $\alpha = -\frac{20}{a - b}$.Substituting this value of $\alpha$ into $(i)$:${\left( -\frac{20}{a - b} ight)^2} + a\left( -\frac{20}{a - b} ight) + 10 = 0$$\frac{400}{(a - b)^2} - \frac{20a}{a - b} + 10 = 0$Multiplying by $(a - b)^2$ (assuming $a 
eq b$):$400 - 20a(a - b) + 10(a - b)^2 = 0$Divide by $10$:$40 - 2a(a - b) + (a - b)^2 = 0$$40 - 2a^2 + 2ab + a^2 - 2ab + b^2 = 0$$40 - a^2 + b^2 = 0$Therefore,$a^2 - b^2 = 40$.

If ${x^2} + ax + 10 = 0$ and ${x^2} + bx - 10 = 0$ have a common root,then ${a^2} - {b^2}$ is equal to

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