
How do you solve \[{x^4} - 5{x^2} \leqslant - 4\] ?
Answer
521.7k+ views
Hint: The equation is an algebraic equation, where the algebraic equation is the combination of constants and variables. To solve the above algebraic equation, we use the tables of multiplication and we can find the value of x
Complete step by step solution:
The algebraic expression is an expression which consists of variables and is consistent with the arithmetic operations. The above equation is a linear equation where the linear equation is defined as the equations are of the first order. These equations are defined for lines in the coordinate system. To solve this linear equation, we apply simple methods. Since by solving these types of equations we get only one value.
Now we solve the given equation, let us consider the equation
\[{x^4} - 5{x^2} \leqslant - 4\]
In LHS we can take \[{x^2}\] common and the equation is written as
\[ \Rightarrow {x^2}({x^2} - 5) \leqslant - 4\]
So we have
\[ \Rightarrow {x^2} \leqslant - 4\,\,\,\,\,and\,\,\,\,({x^2} - 5) \leqslant - 4\]
Let we consider the first inequality we have
\[ \Rightarrow {x^2} \leqslant - 4\]
Taking the square root on both sides we have
\[ \Rightarrow x \leqslant + 2i\,\,\,and\,\,\,x \leqslant - 2i\] where \[\sqrt { - 1} = i\]
Let we consider the second inequality we have
\[ \Rightarrow {x^2} - 5 \leqslant - 4\]
Take -5 to the RHS we get
\[ \Rightarrow {x^2} \leqslant - 4 + 5\]
On simplifying we have
\[ \Rightarrow {x^2} \leqslant 1\]
Taking the square root on both sides we have
\[ \Rightarrow x \leqslant + 1\,\,and\,\,x \leqslant - 1\]
Therefore the values of x are less than or equal to \[ + 2i, - 2i,1\,and\, - 1\]
So, the correct answer is “ \[ + 2i, - 2i,1\,and\, - 1\] ”.
Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. We have 3 types of algebraic expressions namely monomial expression, binomial expression and polynomial expression. By using the tables of multiplication, we can solve the equation.
Complete step by step solution:
The algebraic expression is an expression which consists of variables and is consistent with the arithmetic operations. The above equation is a linear equation where the linear equation is defined as the equations are of the first order. These equations are defined for lines in the coordinate system. To solve this linear equation, we apply simple methods. Since by solving these types of equations we get only one value.
Now we solve the given equation, let us consider the equation
\[{x^4} - 5{x^2} \leqslant - 4\]
In LHS we can take \[{x^2}\] common and the equation is written as
\[ \Rightarrow {x^2}({x^2} - 5) \leqslant - 4\]
So we have
\[ \Rightarrow {x^2} \leqslant - 4\,\,\,\,\,and\,\,\,\,({x^2} - 5) \leqslant - 4\]
Let we consider the first inequality we have
\[ \Rightarrow {x^2} \leqslant - 4\]
Taking the square root on both sides we have
\[ \Rightarrow x \leqslant + 2i\,\,\,and\,\,\,x \leqslant - 2i\] where \[\sqrt { - 1} = i\]
Let we consider the second inequality we have
\[ \Rightarrow {x^2} - 5 \leqslant - 4\]
Take -5 to the RHS we get
\[ \Rightarrow {x^2} \leqslant - 4 + 5\]
On simplifying we have
\[ \Rightarrow {x^2} \leqslant 1\]
Taking the square root on both sides we have
\[ \Rightarrow x \leqslant + 1\,\,and\,\,x \leqslant - 1\]
Therefore the values of x are less than or equal to \[ + 2i, - 2i,1\,and\, - 1\]
So, the correct answer is “ \[ + 2i, - 2i,1\,and\, - 1\] ”.
Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. We have 3 types of algebraic expressions namely monomial expression, binomial expression and polynomial expression. By using the tables of multiplication, we can solve the equation.
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