Let the equation x(x+2)(12-k)=2 have equal ro | vedclass

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(A) Given the equation: $x(x+2)(12-k)=2$. Let $\lambda = 12-k$. Since the equation has equal roots,$k 
eq 12$,so $\lambda 
eq 0$. The equation becom

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Given the equation: $x(x+2)(12-k)=2$. Let $\lambda = 12-k$. Since the equation has equal roots,$k 
eq 12$,so $\lambda 
eq 0$. The equation becomes $\lambda(x^2+2x) = 2$,or $\lambda x^2 + 2\lambda x - 2 = 0$. For equal roots,the discriminant $D = b^2 - 4ac = 0$. $(2\lambda)^2 - 4(\lambda)(-2) = 0$. $4\lambda^2 + 8\lambda = 0$. $4\lambda(\lambda + 2) = 0$. Since $\lambda 
eq 0$,we have $\lambda = -2$. Substituting back,$12-k = -2$,which gives $k = 14$. The point is $(k, \frac{k}{2}) = (14, \frac{14}{2}) = (14, 7)$. The distance $d$ from the point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$. $d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2+4^2}} = \frac{|42 + 28 + 5|}{\sqrt{9+16}} = \frac{75}{5} = 15$.

Let the equation $x(x+2)(12-k)=2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3x+4y+5=0$ is

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