The coefficients a, b and c of the quadratic | vedclass

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Question

(B) The quadratic equation is $ax^{2} + bx + c = 0$. For the equation to have equal roots,the discriminant $D$ must be zero,i.e.,$D = b^{2} - 4ac = 0$

Vedclass · Accepted Answer

$\frac{5}{216}$

Vedclass · Answer

(B) The quadratic equation is $ax^{2} + bx + c = 0$. For the equation to have equal roots,the discriminant $D$ must be zero,i.e.,$D = b^{2} - 4ac = 0$,which implies $b^{2} = 4ac$. Since $a, b, c$ are obtained by throwing a die three times,each variable can take values from the set ${1, 2, 3, 4, 5, 6}$. The total number of outcomes is $6 	imes 6 	imes 6 = 216$. We need to find the number of triplets $(a, b, c)$ such that $b^{2} = 4ac$: $1$. If $a=1, c=1$,then $b^{2} = 4(1)(1) = 4 \Rightarrow b=2$. Triplets: $(1, 2, 1)$. $2$. If $a=1, c=4$,then $b^{2} = 4(1)(4) = 16 \Rightarrow b=4$. Triplets: $(1, 4, 4)$. $3$. If $a=4, c=1$,then $b^{2} = 4(4)(1) = 16 \Rightarrow b=4$. Triplets: $(4, 4, 1)$. $4$. If $a=2, c=2$,then $b^{2} = 4(2)(2) = 16 \Rightarrow b=4$. Triplets: $(2, 4, 2)$. $5$. If $a=3, c=3$,then $b^{2} = 4(3)(3) = 36 \Rightarrow b=6$. Triplets: $(3, 6, 3)$. Total favorable outcomes = $5$. Therefore,the required probability is $\frac{5}{216}$.

List-$I$	List-$II$
$A$. Probability of getting $1$ red, $1$ white and $1$ blue ball	$(i)$ $\frac{3}{44}$
$B$. Probability of getting $2$ white and $1$ blue ball	$(ii)$ $\frac{21}{55}$
$C$. Probability of getting $2$ red and $1$ white ball	$(iii)$ $\frac{38}{55}$
$D$. Probability that none of the balls is white	$(iv)$ $\frac{3}{11}$
	$(v)$ $\frac{9}{110}$

The coefficients $a, b$ and $c$ of the quadratic equation $ax^{2} + bx + c = 0$ are obtained by throwing a die three times. The probability that this equation has equal roots is

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