Let a ne b, c ne 0. If the equations x^2 + ax | vedclass

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(D) Let the roots of the equations be:$x^2 + ax + bc = 0 \rightarrow \alpha, \beta$  $(1)$$x^2 + bx + ac = 0 \rightarrow \alpha, \gamma$  $(2)$Subtrac

Vedclass · Accepted Answer

Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is the correct explanation of Statement $-1$.

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(D) Let the roots of the equations be:$x^2 + ax + bc = 0 ightarrow \alpha, \beta$  $(1)$$x^2 + bx + ac = 0 ightarrow \alpha, \gamma$  $(2)$Subtracting $(2)$ from $(1)$:$(a - b)x + (bc - ac) = 0$$(a - b)x - c(a - b) = 0$Since $a 
e b$,we divide by $(a - b)$ to get $x = c$.Thus,the common root is $\alpha = c$.Substituting $x = c$ into $(1)$:$c^2 + ac + bc = 0 \Rightarrow c(c + a + b) = 0$.Since $c 
e 0$,we have $a + b + c = 0$.From the product of roots in $(1)$,$\alpha \cdot \beta = bc \Rightarrow c \cdot \beta = bc \Rightarrow \beta = b$.From the product of roots in $(2)$,$\alpha \cdot \gamma = ac \Rightarrow c \cdot \gamma = ac \Rightarrow \gamma = a$.The equation with roots $\beta$ and $\gamma$ is $x^2 - (\beta + \gamma)x + \beta \gamma = 0$.$x^2 - (b + a)x + ab = 0$.Since $a + b = -c$,the equation becomes $x^2 - (-c)x + ab = 0$,which is $x^2 + cx + ab = 0$.Both statements are true,and Statement $-2$ explains Statement $-1$.

Let $a \ne b, c \ne 0$. If the equations $x^2 + ax + bc = 0$ and $x^2 + bx + ac = 0$ have a common root,then:
Statement $-1$: The equation of the other roots is $x^2 + cx + ab = 0$.
Statement $-2$: $a + b + c = 0$.

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Let $a \ne b, c \ne 0$. If the equations $x^2 + ax + bc = 0$ and $x^2 + bx + ac = 0$ have a common root,then:Statement $-1$: The equation of the other roots is $x^2 + cx + ab = 0$.Statement $-2$: $a + b + c = 0$.