The product of two alternate odd integers exceeds three times the smaller by $12$ . What is the larger number?

3

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Solution

The correct option is B $6$
Let the smaller odd integer be 'a' then
other be $(a + 2)$
$a (a + 2) = 3 a + 12 \Rightarrow a^{2} + 2 a = 3 a + 12$
$\Rightarrow a^{2} - a - 12 = 0 \Rightarrow a = 4_{n} - 3$
$∴$ larger no $6$

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The product of two alternate odd integers exceeds three times the smaller by 12. What is the larger number?

The correct option is B 6Let the smaller odd integer be 'a' thenother be (a+2)a(a+2)=3a+12⇒a2+2a=3a+12⇒a2−a−12=0⇒a=4n−3∴ larger no 6

The product of two alternate odd integers exceeds three times the smaller by $12$ . What is the larger number?

The correct option is B $6$
Let the smaller odd integer be 'a' then
other be $(a + 2)$
$a (a + 2) = 3 a + 12 \Rightarrow a^{2} + 2 a = 3 a + 12$
$\Rightarrow a^{2} - a - 12 = 0 \Rightarrow a = 4_{n} - 3$
$∴$ larger no $6$