Question

If $\left| a \right| = 3$, $\left| b \right| = 4$ and $\left| {a + b} \right| = 5$ then $\left| {a - b} \right|$ is equal to?
A) 6
B) 5
C) 4
D) 3

Answer 1

Here in this question based on the algebraic equation is a combination of constant and variables. we have to find the value of $\left| {a - b} \right|$ using given conditions or values. since the equation involves the modulus, by using the definition of modulus or absolute value and further simplifying by using algebraic identities and simple arithmetic operation to get the required solution.

Complete step by step answer:
The absolute value or modulus of a real function $f\left( x \right)$, it is denoted as $\left| {f\left( x \right)} \right|$, is the non-
negative value of $f\left( x \right)$ without considering its sign. The value of $\left| {f\left( x \right)} \right|$ defined as

{ + f\left( x \right);\,\,\,\,f\left( x \right) \geqslant 0} \\\ { - f\left( x \right);\,\,\,\,f\left( x \right) < 0} \end{array}} \right.$$ Consider the question: Given, the value of $$\left| a \right| = 3$$, $$\left| b \right| = 4$$ and $$\left| {a + b} \right| = 5$$ We have to find the value of $$\left| {a - b} \right| = ?$$ Now, let us consider a expression: $$ \Rightarrow \,\,\,{\left| {a + b} \right|^2} + {\left| {a - b} \right|^2}$$ ------(1) We know that the algebraic identities like $${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$$ and $${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$$ On applying the identities, equation (1) becomes $$ \Rightarrow \,\,\,{\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right| + {\left| a \right|^2} + {\left| b \right|^2} - 2\left| a \right|\left| b \right|$$ On simplification, we get $$ \Rightarrow \,\,\,{\left| a \right|^2} + {\left| b \right|^2} + {\left| a \right|^2} + {\left| b \right|^2}$$ $$ \Rightarrow \,\,\,2{\left| a \right|^2} + 2{\left| b \right|^2}$$ Take 2 as common, then we have $$ \Rightarrow \,\,\,2\left( {{{\left| a \right|}^2} + {{\left| b \right|}^2}} \right)$$ $$\therefore \,\,\,\,{\left| {a + b} \right|^2} + {\left| {a - b} \right|^2} = 2\left( {{{\left| a \right|}^2} + {{\left| b \right|}^2}} \right)$$ On substituting the given values, then $$ \Rightarrow \,\,\,\,\,\,{\left| 5 \right|^2} + {\left| {a - b} \right|^2} = 2\left( {{{\left| 3 \right|}^2} + {{\left| 4 \right|}^2}} \right)$$ $$ \Rightarrow \,\,\,\,\,\,{5^2} + {\left| {a - b} \right|^2} = 2\left( {{3^2} + {4^2}} \right)$$ $$ \Rightarrow \,\,\,\,\,\,25 + {\left| {a - b} \right|^2} = 2\left( {9 + 16} \right)$$ On simplification, we get $$ \Rightarrow \,\,\,\,\,\,25 + {\left| {a - b} \right|^2} = 2\left( {25} \right)$$ $$ \Rightarrow \,\,\,\,\,\,25 + {\left| {a - b} \right|^2} = 50$$ Subtract 25 on both side, then we have $$ \Rightarrow \,\,\,\,\,\,{\left| {a - b} \right|^2} = 50 - 25$$ $$ \Rightarrow \,\,\,\,\,\,{\left| {a - b} \right|^2} = 25$$ Take square root on both the sides, then $$ \Rightarrow \,\,\,\,\,\,\left| {a - b} \right| = \pm \sqrt {25} $$ As we know 25 is the square number of 5 i.e., $${5^2} = 25$$ $$ \Rightarrow \,\,\,\,\,\,\left| {a - b} \right| = \pm \sqrt {{5^2}} $$ On cancelling the square and root in RHS, then $$ \Rightarrow \,\,\,\,\,\,\left| {a - b} \right| = \pm \,5$$ Here, we have to choose only positive values because $$\left| {a - b} \right|$$ is an absolute value. $$\therefore \,\,\,\,\left| {a - b} \right| = 5$$ Hence, the required solution $$\left| {a - b} \right| = 5$$ **Therefore, option (B) is the correct answer.** **Note:** The absolute value or modulus of a real number $$x$$, denoted $$\left| x \right|$$, is the non-negative value of $$x$$ without regard to its sign i.e., $$\left| x \right| = x$$. Students should remember the basic algebraic identities on using the identities; the solution becomes easier to solve and know the square numbers of at least 1-100.