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

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(D) Let the common root be $\alpha$. Then:$\alpha^2 + a\alpha + bc = 0$  ....$(1)$$\alpha^2 + b\alpha + ac = 0$  ....$(2)$Subtracting $(2)$ from $(1)$

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 common root be $\alpha$. Then:$\alpha^2 + a\alpha + bc = 0$  ....$(1)$$\alpha^2 + b\alpha + ac = 0$  ....$(2)$Subtracting $(2)$ from $(1)$:$(a - b)\alpha + bc - ac = 0$$(a - b)\alpha - c(a - b) = 0$Since $a 
e b$,we divide by $(a - b)$ to get $\alpha = c$.Substituting $\alpha = c$ into $(1)$:$c^2 + ac + bc = 0$$c(c + a + b) = 0$Since $c 
e 0$,we have $a + b + c = 0$. This proves Statement-$2$ is true.Now,let the roots of the first equation be $c$ and $\beta$. From the product of roots,$c\beta = bc \implies \beta = b$.Let the roots of the second equation be $c$ and $\gamma$. From the product of roots,$c\gamma = ac \implies \gamma = a$.The equation with roots $\beta = b$ and $\gamma = a$ is:$x^2 - (a + b)x + ab = 0$Since $a + b = -c$,the equation becomes $x^2 - (-c)x + ab = 0$,which is $x^2 + cx + ab = 0$. This proves Statement-$1$ is true.Since Statement-$2$ $(a+b+c=0)$ is used to derive the form of the equation in Statement-$1$,Statement-$2$ is the correct explanation of 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-$2$: $a + b + c = 0$.