
How do you solve the equation \[5{\left( {x - 4} \right)^2} = 125\] ?
Answer
496.8k+ views
Hint: To solve this equation by using completing the square, or complete the square, is a method that can be used to solve quadratic equation, first rearrange the given equation in the form \[{\left( {x + k} \right)^2} + A = 0\] where \[x\] is the variable and k and A are constants. Then by the resultant equation on simplification we get the required root or value of variable \[x\] .
Complete step-by-step answer:
Completing the square method is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a quadratic equation. ‘Quad’ means four but ‘Quadratic’ means ‘to make square’. A quadratic equation in its standard form is represented as:
\[a{x^2} + bx + c = 0\] , where a, b and c are real numbers such that \[a \ne 0\] and \[x\] is a variable
Since the degree of the equation \[a{x^2} + bx + c = 0\] is two; it will have two roots or solutions. The roots of polynomials are the values of \[x\] which satisfy the equation. There are several methods to find the roots of a quadratic equation. One of them is by completing the square.
Consider the given equation
\[ \Rightarrow 5{\left( {x - 4} \right)^2} = 125\]
Divide both side by 5, then
\[ \Rightarrow \dfrac{5}{5}{\left( {x - 4} \right)^2} = \dfrac{{125}}{5}\]
\[ \Rightarrow \dfrac{5}{5}{\left( {x - 4} \right)^2} = \dfrac{{125}}{5}\]
\[ \Rightarrow {\left( {x - 4} \right)^2} = 25\]
Take square root on both side, then
\[ \Rightarrow \,\,\,\sqrt {\,{{\left( {x - 4} \right)}^2}} = \pm \sqrt {25} \]
25 is the square root of 5 i.e., \[\sqrt {25} = \sqrt {{5^2}} = 5\]
\[ \Rightarrow \left( {x - 4} \right) = \pm 5\]
Add 4 on both side, then
\[ \Rightarrow x - 4 + 4 = \pm 5 + 4\]
\[ \Rightarrow x = \pm 5 + 4\]
\[ \Rightarrow x = + 5 + 4\] or \[x = - 5 + 4\]
\[ \Rightarrow x = 9\] or \[x = - 1\]
Hence, the value of the variable \[x\] in the equation \[5{\left( {x - 4} \right)^2} = 125\] is \[x = 9\] or \[x = - 1\] .
So, the correct answer is “ \[x = 9\] or \[x = - 1\] ”.
Note: The given equation is a form of quadratic equation. To solve this we are not following the general method to find the value of variable x. We are using the concept of square and square root concept to simplify the given equation and for the further simplification we are going to use the arithmetic operations.
Complete step-by-step answer:
Completing the square method is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a quadratic equation. ‘Quad’ means four but ‘Quadratic’ means ‘to make square’. A quadratic equation in its standard form is represented as:
\[a{x^2} + bx + c = 0\] , where a, b and c are real numbers such that \[a \ne 0\] and \[x\] is a variable
Since the degree of the equation \[a{x^2} + bx + c = 0\] is two; it will have two roots or solutions. The roots of polynomials are the values of \[x\] which satisfy the equation. There are several methods to find the roots of a quadratic equation. One of them is by completing the square.
Consider the given equation
\[ \Rightarrow 5{\left( {x - 4} \right)^2} = 125\]
Divide both side by 5, then
\[ \Rightarrow \dfrac{5}{5}{\left( {x - 4} \right)^2} = \dfrac{{125}}{5}\]
\[ \Rightarrow \dfrac{5}{5}{\left( {x - 4} \right)^2} = \dfrac{{125}}{5}\]
\[ \Rightarrow {\left( {x - 4} \right)^2} = 25\]
Take square root on both side, then
\[ \Rightarrow \,\,\,\sqrt {\,{{\left( {x - 4} \right)}^2}} = \pm \sqrt {25} \]
25 is the square root of 5 i.e., \[\sqrt {25} = \sqrt {{5^2}} = 5\]
\[ \Rightarrow \left( {x - 4} \right) = \pm 5\]
Add 4 on both side, then
\[ \Rightarrow x - 4 + 4 = \pm 5 + 4\]
\[ \Rightarrow x = \pm 5 + 4\]
\[ \Rightarrow x = + 5 + 4\] or \[x = - 5 + 4\]
\[ \Rightarrow x = 9\] or \[x = - 1\]
Hence, the value of the variable \[x\] in the equation \[5{\left( {x - 4} \right)^2} = 125\] is \[x = 9\] or \[x = - 1\] .
So, the correct answer is “ \[x = 9\] or \[x = - 1\] ”.
Note: The given equation is a form of quadratic equation. To solve this we are not following the general method to find the value of variable x. We are using the concept of square and square root concept to simplify the given equation and for the further simplification we are going to use the arithmetic operations.
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