Question

The product of two consecutive even number is 224. Then the numbers are:

Answer 1

${\large{\sf{\pmb{\underline{\maltese \: \: Undertanding \; the \; Question...}}}}}$

• This question says that we have to find the numbers whose product of two consecutive even number is 224. This question is from very intersting topic of mathematics named Linear equations. According to the question's answer we have to use Quadratic equations too here as factorising. Let's solve this question right now!

${\large{\sf{\pmb{\underline{\maltese \: \: Given \; that}}}}}$

• The product of two consecutive even number is 224.

${\large{\sf{\pmb{\underline{\maltese \: \: To \; find}}}}}$

• The original number.

${\large{\sf{\pmb{\underline{\maltese \: \: Solution}}}}}$

• The original number = 14 or -16

${\large{\sf{\pmb{\underline{\maltese \: \: Using \; concepts}}}}}$

• Linear equation.
• Quadratic equation.

${\large{\sf{\pmb{\underline{\maltese \: \: Assumptions}}}}}$

• Let first number is a
• Let second number is (a+2)

${\large{\sf{\pmb{\underline{\maltese \: \: Knowledge \; required}}}}}$

→ Factorisation

When we going to factorise an algebraic expression then we have to write it's factorised products. These factors may be number, algebraic variable or algebraic expression. For example,

→ x²-5x-6=0

→ x²-6x+x-6 = 0

→ x(x-6) + 1(x-6) = 0

→ (x+1)(x-6) = 0

→ x = -1 or x = 6

→ x = -1 or +6

There are many ways to factorise. But the most important way is middle term splitting method.

→ Factorised identities

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$

→ Linear equation = Here we will have to deal with linear expressions in just one variable. Such equations are known to be “linear equation in one variable”

→ An algebraic equation in an equality involving variable. It has an equality sign. The expression on the left of equality sign is LHS. The expression on the right of equality sign is RHS like in expression 2x - 3 = 7

${\large{\sf{\pmb{\underline{\maltese \: \: Full \; Solution}}}}}$

↝ First number is a

↝ Second number is (a+2)

Now according to the question,

↝ a(a+2) = 224

↝ a² + 2a - 224 = 0

↝ a² + 16a - 14a - 224 = 0

↝ a(a+16) - 14(a+16) = 0

↝ (a-14) (a+16)

↝ a = 14, a = -16

• Henceforth, the numbers are 14 or -16.