If 2x^2+3x-2=0 and 3x^2+ax-2=0 have one commo | vedclass

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(B) First,solve the quadratic equation $2x^2+3x-2=0$ by factoring: $2x^2+4x-x-2=0$ $\Rightarrow 2x(x+2)-1(x+2)=0$ $\Rightarrow (2x-1)(x+2)=0$. Thus,th

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$7$

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(B) First,solve the quadratic equation $2x^2+3x-2=0$ by factoring: $2x^2+4x-x-2=0$ $\Rightarrow 2x(x+2)-1(x+2)=0$ $\Rightarrow (2x-1)(x+2)=0$. Thus,the roots are $x=-2$ and $x=\frac{1}{2}$. Case $1$: If $x=-2$ is the common root,substitute it into $3x^2+ax-2=0$: $3(-2)^2+a(-2)-2=0$ $\Rightarrow 12-2a-2=0$ $\Rightarrow 10=2a$ $\Rightarrow a=5$. Case $2$: If $x=\frac{1}{2}$ is the common root,substitute it into $3x^2+ax-2=0$: $3(\frac{1}{2})^2+a(\frac{1}{2})-2=0$ $\Rightarrow \frac{3}{4}+\frac{a}{2}-2=0$ $\Rightarrow \frac{a}{2} = 2 - \frac{3}{4} = \frac{5}{4}$ $\Rightarrow a=2.5$. The sum of all possible values of $a$ is $5+2.5=7.5$.

If $2x^2+3x-2=0$ and $3x^2+ax-2=0$ have one common root,then the sum of all possible values of $a$ is (in $.5$)

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