Question

Given $m = - 2$ and $n = 6$ Evaluate,
$(i)m + n$
$(ii)2{m^2} + 8n$
$(iii){m^2} - {n^2}$
$(iv)3{m^3} + 2{m^2}n + 3{n^3}$

Answer 1

Hint: First, from the given that there are two different values $m = - 2$ and $n = 6$.
From these two values, we have to find the values given in the several options in the quadratic equation, cubic equation, and linear degree equation.
A linear equation is the degree one equation which is the straight line.
Quadratic is the at most two-degree equation and cubic is at the most three-degree equation.

Complete step by step answer:
$(i)m + n$
First, solve option one with $m = - 2$ and $n = 6$.
The addition is the sum of adding two or more than two numbers as variables.
Thus, we get while solving this $m + n = - 2 + 6$(if the positive term is greater than the resultant will be in positive)
Further solving the equation, we get, $m + n = 4$

$(ii)2{m^2} + 8n$
Since the general quadratic equation can be written as in the form of \[a{m^2} - bm + c = 0\] where a is the quadratic root with at most two-degree it will be not represented as any terms, b is the sum of the given terms, c is the product of the terms.
But here we just need to apply the given values in the equation that $2{m^2} + 8n$
Substitute $m = - 2$ and $n = 6$ we get \[2{m^2} + 8n = 2{( - 2)^2} + 8(6)\] (again by the help of adding and multiplication)
We get after solving the equation as \[2{m^2} + 8n = 2 \times 4 + 48 \Rightarrow 56\]

$(iii){m^2} - {n^2}$
similarly, if the degree two-equation as above,
Hence solving this from $m = - 2$ and $n = 6$, we get after applying the values ${m^2} - {n^2} = {( - 2)^2} - {(6)^2}$
We will see what multiplication is, Multiplicand refers to the number multiplied. The multiplier is the number that refers to the number which multiplies the first number.
Using the information, we used the square in multiplication concept, thus we get ${m^2} - {n^2} = 4 - 36 \Rightarrow - 32$
subtraction which is the minus of two or more than two numbers or values but here comes the condition that in subtraction the greater number sign will stay constant.

$(iv)3{m^3} + 2{m^2}n + 3{n^3}$
This is the degree three equation and known as the cubic equation.
Put $m = - 2$and$n = 6$we get, $3{m^3} + 2{m^2}n + 3{n^3} = 3{( - 2)^3} + 2{( - 2)^2}(6) + 3{(6)^3}$
Cubic root means $2{}^3 = 2 \times 2 \times 2 = 8$from this we can obtain the above solution as,$3{m^3} + 2{m^2}n + 3{n^3} = 3 \times ( - 8) + 2 \times 4 \times (6) + 3 \times 216$
Further solving with the help of multiplication, addition, subtraction we get
$3{m^3} + 2{m^2}n + 3{n^3} = - 24 + 48 + 648 \Rightarrow 672$

Note: Subtraction the greater number sign will stay constant example $2 - 3 = - 1$ three has the greater sign with negative thus the resultant answer is negative if suppose three is positive and two is negative then the resultant answer will be positive.
There is another operation that is not seen in the question $\dfrac{m}{n} = \dfrac{{ - 2}}{6} = \dfrac{{ - 1}}{3} = - 0.333...$ which is the division of a given problem.