Question

If $ {x^{{{(5 + a)}^2}}} \times {x^{{{(5 - a)}^2}}} = {x^{40}} $ then find a.

Answer 1

Hint: Here we will use additive law for the power and exponent and then simplify the equations using the square of the whole square. Also, we will use the square and square-root concepts and then simplify for the required solution.

Complete step-by-step answer:
Take the given expression –
 $ {x^{{{(5 + a)}^2}}} \times {x^{{{(5 - a)}^2}}} = {x^{40}} $
By using the law of power and exponent which states that when bases are the same powers are added in case we have multiplication sign in between two powers and exponents. Such as $ {x^m} \times {x^n} = {x^{m + n}} $
 $ {x^{{{(5 + a)}^2} + }}^{{{(5 - a)}^2}} = {x^{40}} $
When bases are the same on both the sides of the equation, then the powers are equal.
 $ \Rightarrow {(5 + a)^2} + {(5 - a)^2} = 40 $
Use the identity for the whole square of two terms. $ {(a + b)^2} = {a^2} + 2ab + {b^2} $ and $ {(a - b)^2} = {a^2} - 2ab + {b^2} $
 $ \Rightarrow \left( {{5^2} + 2(5)(a) + {a^2}} \right) + \left( {{5^2} - 2(5)(a) + {a^2}} \right) = 40 $
Simplify the above equation-
 $ \Rightarrow \left( {25 + 10a + {a^2}} \right) + \left( {25 - 10a + {a^2}} \right) = 40 $
Open the brackets. Remember when there is a positive sign outside the bracket then the sign of the terms inside the brackets do not change.
 $ \Rightarrow 25 + 10a + {a^2} + 25 - 10a + {a^2} = 40 $
Make the pair of like terms in the above equation.
 $ \Rightarrow \underline {25 + 25} + \underline {10a - 10a} + {\underline {{a^2} + a} ^2} = 40 $
Do simplification in the above equation. When the like terms are with equal value and opposite signs cancel each other.
 $ \Rightarrow 50 + 2{a^2} = 40 $
Move constantly on the right side. Remember when any term is moved from one side to another then the sign is also changed. Positive term becomes negative and negative term becomes positive.
 $ \Rightarrow 2{a^2} = 40 - 50 $
Simplify the above equation –
 $ \Rightarrow 2{a^2} = - 10 $
When any term is multiplicative on one side is moved to another side, then it goes to the denominator.
 $ \Rightarrow {a^2} = \dfrac{{ - 10}}{2} $
Find the multiples in the numerator on the right side of the equation.
 $ \Rightarrow {a^2} = \dfrac{{ - 5 \times 2}}{2} $
Common multiples from the numerator and the denominator cancel each other.
 $ \Rightarrow {a^2} = ( - 5) $
Take square-root on both the sides of the equation –
 $ \Rightarrow \sqrt {{a^2}} = \sqrt {( - 5)} $
Square and square-root cancel each other on the right hand side of the equation.
 $ \Rightarrow a = \sqrt {( - 5)} $
This is the required solution.
So, the correct answer is “ $ a = \sqrt {( - 5)} $ ”.

Note: Always remember when there is a negative sign outside the bracket then sign of the terms inside the bracket changes, negative term becomes positive and vice-versa and when there is positive sign outside the bracket then there is no change in the sign of the terms inside of the bracket.