Let the equations ax^2-7x+c=0 and ax^2+5x-c=0 | vedclass

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(B) Let the common root be $\alpha$. Given equations are $ax^2-7x+c=0$ and $ax^2+5x-c=0$. Since $\alpha$ is a common root,we have: $a\alpha^2-7\alpha+

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$\frac{1}{2}$

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(B) Let the common root be $\alpha$. Given equations are $ax^2-7x+c=0$ and $ax^2+5x-c=0$. Since $\alpha$ is a common root,we have: $a\alpha^2-7\alpha+c=0$ and $a\alpha^2+5\alpha-c=0$. Subtracting the two equations: $(ax^2-7x+c) - (ax^2+5x-c) = 0$ $\Rightarrow -12x + 2c = 0$ $\Rightarrow x = \frac{c}{6}$. Thus,the common root $\alpha = \frac{c}{6}$. Substituting $\alpha = \frac{c}{6}$ into $ax^2-7x+c=0$: $a(\frac{c}{6})^2 - 7(\frac{c}{6}) + c = 0 \Rightarrow \frac{ac^2}{36} - \frac{7c}{6} + c = 0$. Since $c 
eq 0$,divide by $c$: $\frac{ac}{36} - \frac{7}{6} + 1 = 0$ $\Rightarrow \frac{ac}{36} = \frac{1}{6}$ $\Rightarrow ac = 6$. Given that $3$ is a root of $ax^2-7x+c=0$ other than the common root: $a(3)^2 - 7(3) + c = 0$ $\Rightarrow 9a - 21 + c = 0$ $\Rightarrow c = 21 - 9a$. Substitute $c = 21 - 9a$ into $ac = 6$: $a(21 - 9a) = 6$ $\Rightarrow 21a - 9a^2 = 6$ $\Rightarrow 3a^2 - 7a + 2 = 0$. Solving for $a$: $(3a-1)(a-2) = 0$,so $a = 2$ or $a = \frac{1}{3}$. If $a = 2$,then $c = 21 - 9(2) = 3$. The equation $ax^2-7x+c=0$ becomes $2x^2-7x+3=0$. Factoring: $(2x-1)(x-3) = 0$,so roots are $3$ and $\frac{1}{2}$. The common root is $\frac{1}{2}$.

Let the equations $ax^2-7x+c=0$ and $ax^2+5x-c=0$ have a common root and $ac \neq 0$. If $3$ is a root of $ax^2-7x+c=0$ other than the common root,then the common root of the given equations is

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