
How do you solve \[4{x^2} = 16x\]?
Answer
548.1k+ views
Hint: According to the above question, we will first try to rearrange the equation in such a way that all the like terms are placed on one side. Then we will factorize the equation by taking out the common factor. This makes it easy to solve, and we will get the answer.
Complete step by step solution:
The given question is to solve the equation:
\[4{x^2} = 16x\]
First, we will try to rearrange the equation. We will try to keep the like terms on one side. We will add \[ - 16x\]on both the sides, and then we will get:
\[ \Rightarrow 4{x^2} - 16x = 16x - 16x\]
\[ \Rightarrow 4{x^2} - 16x = 0\]
Now, we will take out all the common factors from the equation. Here the common factor will be \[4x\]. We will take out \[4x\]from the equation on one side, and then we will get:
\[ \Rightarrow 4x(x - 4) = 0\]
Now, we will try to again rewrite or rearrange the equation. We will try to separate \[4x\]in the equation, and we will get:
\[ \Rightarrow 4(x - 4) \cdot x = 0\]
Now, we will separate the equations and try to solve it. Here we will solve the equations with the help of the null law method. In this method we need to set the first group and set it equal to 0 and then solve. For solving algebraic or quadratic equations, this law can be used. In this law, a mathematical expression is used to be converted into an equation by making the equation equal to 0, and it becomes easy to get the answer after that:
\[ \Rightarrow 4(x - 4) = 0\,\,and\,\,x = 0\]
These are the two equations. We will solve the first equation:
\[4(x - 4) = 0\]
Now, we will divide both the sides by 4 to get rid of 4 from the left side, and we will get:
\[ \Rightarrow \dfrac{{4(x - 4)}}{4} = \dfrac{0}{4}\]
\[ \Rightarrow (x - 4) = 0\]
Now, we will add 4 on both the sides, and we will get:
\[ \Rightarrow x - 4 + 4 = 0 + 4\]
Now, by solving the equation, we get:
\[ \Rightarrow x = 4\]
Therefore, the values of ‘x’ from both the equations are 4,0.
Note: Factorisation is a method in which a mathematical expression or a polynomial is divided into different terms so that it becomes easier to solve the expression. In the factorization method, we reduce any algebraic or quadratic equation into its simpler form.
Complete step by step solution:
The given question is to solve the equation:
\[4{x^2} = 16x\]
First, we will try to rearrange the equation. We will try to keep the like terms on one side. We will add \[ - 16x\]on both the sides, and then we will get:
\[ \Rightarrow 4{x^2} - 16x = 16x - 16x\]
\[ \Rightarrow 4{x^2} - 16x = 0\]
Now, we will take out all the common factors from the equation. Here the common factor will be \[4x\]. We will take out \[4x\]from the equation on one side, and then we will get:
\[ \Rightarrow 4x(x - 4) = 0\]
Now, we will try to again rewrite or rearrange the equation. We will try to separate \[4x\]in the equation, and we will get:
\[ \Rightarrow 4(x - 4) \cdot x = 0\]
Now, we will separate the equations and try to solve it. Here we will solve the equations with the help of the null law method. In this method we need to set the first group and set it equal to 0 and then solve. For solving algebraic or quadratic equations, this law can be used. In this law, a mathematical expression is used to be converted into an equation by making the equation equal to 0, and it becomes easy to get the answer after that:
\[ \Rightarrow 4(x - 4) = 0\,\,and\,\,x = 0\]
These are the two equations. We will solve the first equation:
\[4(x - 4) = 0\]
Now, we will divide both the sides by 4 to get rid of 4 from the left side, and we will get:
\[ \Rightarrow \dfrac{{4(x - 4)}}{4} = \dfrac{0}{4}\]
\[ \Rightarrow (x - 4) = 0\]
Now, we will add 4 on both the sides, and we will get:
\[ \Rightarrow x - 4 + 4 = 0 + 4\]
Now, by solving the equation, we get:
\[ \Rightarrow x = 4\]
Therefore, the values of ‘x’ from both the equations are 4,0.
Note: Factorisation is a method in which a mathematical expression or a polynomial is divided into different terms so that it becomes easier to solve the expression. In the factorization method, we reduce any algebraic or quadratic equation into its simpler form.
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