The quadratic equations x^2-6x+a=0 and x^2-cx | vedclass

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(C) Let the common root be $\alpha$. The equations are $x^2-6x+a=0$ and $x^2-cx+6=0$.Let the other roots be $\beta_1$ and $\beta_2$ respectively. Give

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the common root be $\alpha$. The equations are $x^2-6x+a=0$ and $x^2-cx+6=0$.Let the other roots be $\beta_1$ and $\beta_2$ respectively. Given $\beta_1 : \beta_2 = 4:3$,so $\beta_1 = 4k$ and $\beta_2 = 3k$ for some constant $k$.From the properties of roots:For the first equation: $\alpha + 4k = 6$ and $\alpha \cdot 4k = a$.For the second equation: $\alpha + 3k = c$ and $\alpha \cdot 3k = 6$.From $\alpha \cdot 3k = 6$,we have $k = \frac{2}{\alpha}$.Substitute $k$ into the first equation: $\alpha + 4(\frac{2}{\alpha}) = 6 \Rightarrow \alpha + \frac{8}{\alpha} = 6$.Multiplying by $\alpha$: $\alpha^2 - 6\alpha + 8 = 0$.Factoring: $(\alpha - 4)(\alpha - 2) = 0$,so $\alpha = 4$ or $\alpha = 2$.If $\alpha = 4$,then $4k = 6 - 4 = 2 \Rightarrow k = 0.5$. Then $\beta_1 = 4(0.5) = 2$ and $\beta_2 = 3(0.5) = 1.5$. Since $\beta_2$ must be an integer,this is rejected.If $\alpha = 2$,then $4k = 6 - 2 = 4 \Rightarrow k = 1$. Then $\beta_1 = 4(1) = 4$ and $\beta_2 = 3(1) = 3$. Both are integers.Thus,the common root is $2$.

The quadratic equations $x^2-6x+a=0$ and $x^2-cx+6=0$ have one root in common. If the other roots of the first and second equations are integers and are in the ratio $4:3$,then their common root is

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