The equations 2x^2+ax-2=0 and x^2+x+2a=0 have | vedclass

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(D) Let $\alpha$ be the common root of the equations $2x^2+ax-2=0$ and $x^2+x+2a=0$. Then $2\alpha^2+a\alpha-2=0$ and $\alpha^2+\alpha+2a=0$. Multiply

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$\frac{-2+\sqrt{22}}{3}$

Vedclass · Answer

(D) Let $\alpha$ be the common root of the equations $2x^2+ax-2=0$ and $x^2+x+2a=0$. Then $2\alpha^2+a\alpha-2=0$ and $\alpha^2+\alpha+2a=0$. Multiplying the second equation by $2$,we get $2\alpha^2+2\alpha+4a=0$. Subtracting this from the first equation: $(a-2)\alpha - 2 - 4a = 0$,so $\alpha = \frac{4a+2}{a-2}$. Substituting $\alpha$ into $x^2+x+2a=0$: $(\frac{4a+2}{a-2})^2 + \frac{4a+2}{a-2} + 2a = 0$. Solving this for $a$ given $a 
eq 0$,we find $a = -3$. Substituting $a = -3$ into the equation $ax^2-4x-2a=0$,we get $-3x^2-4x+6=0$,or $3x^2+4x-6=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{-4 \pm \sqrt{16 - 4(3)(-6)}}{2(3)} = \frac{-4 \pm \sqrt{88}}{6} = \frac{-2 \pm \sqrt{22}}{3}$. Thus,one of the roots is $\frac{-2+\sqrt{22}}{3}$.

The equations $2x^2+ax-2=0$ and $x^2+x+2a=0$ have exactly one common root. If $a \neq 0$,then one of the roots of the equation $ax^2-4x-2a=0$ is

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