JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 12)
The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
$${1 \over {72}}$$
$${5 \over {216}}$$
$${1 \over {36}}$$
$${1 \over {54}}$$
Explanation
ax2 + bx + c = 0
a, b, c $$ \in $$ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
For equal roots, D = 0 $$ \Rightarrow $$ b2 = 4ac
$$ \Rightarrow $$ ac = $${{{b^2}} \over 4}$$
Favourable case :
If b = 2, ac = 1 $$ \Rightarrow $$ a = 1, c = 1
If b = 4, ac = 4 :
a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 $$ \Rightarrow $$ a = 3, c = 3
$$ \therefore $$ Favorable cases = 5
$$ \therefore $$ Required probability = $${5 \over {216}}$$
a, b, c $$ \in $$ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
For equal roots, D = 0 $$ \Rightarrow $$ b2 = 4ac
$$ \Rightarrow $$ ac = $${{{b^2}} \over 4}$$
Favourable case :
If b = 2, ac = 1 $$ \Rightarrow $$ a = 1, c = 1
If b = 4, ac = 4 :
a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 $$ \Rightarrow $$ a = 3, c = 3
$$ \therefore $$ Favorable cases = 5
$$ \therefore $$ Required probability = $${5 \over {216}}$$
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