
How do you solve \[5{x^2} - 10x = 23\] by completing the square method ?
Answer
536.4k+ views
Hint: The given problem requires us to solve an equation using completing the square method. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formulas and by splitting the middle term. Completing the square method involves manipulating the equation to form a perfect square and solving for variable or unknown.
Complete step-by-step solution:
In the given question, we have to solve the given equation \[5{x^2} - 10x = 23\] using the completing the square method.
So, \[5{x^2} - 10x = 23\]
So, for completing the square, we have to make the given equation resemble the whole square of a binomial identity: ${\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right)$. So, we get,
\[ \Rightarrow {\left( {\sqrt 5 x} \right)^2} - 2\left( {\sqrt 5 x} \right)\left( {\sqrt 5 } \right) = 23\]
Adding the square of the term \[\sqrt 5 \] to both sides of the equation so as to exactly match the above mentioned identity.
\[ \Rightarrow {\left( {\sqrt 5 x} \right)^2} - 2\left( {\sqrt 5 x} \right)\left( {\dfrac{3}{{2\sqrt 5 }}} \right) + {\left( {\sqrt 5 } \right)^2} = 23 + {\left( {\sqrt 5 } \right)^2}\]
So, the three terms on left side of the equation resemble the identity ${\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right)$. Hence, \[\left[ {{{\left( {\sqrt 5 x} \right)}^2} - 2\left( {\sqrt 5 x} \right)\left( {\dfrac{3}{{2\sqrt 5 }}} \right) + {{\left( {\sqrt 5 } \right)}^2}} \right]\] can be condensed as \[{\left[ {\left( {\sqrt 5 x} \right) - \left( {\sqrt 5 } \right)} \right]^2}\]. So, we get,
\[ \Rightarrow {\left( {\sqrt 5 x - \sqrt 5 } \right)^2} = 28\]
Taking out \[5\] from inside the square term,
\[ \Rightarrow 5{\left( {x - 1} \right)^2} = 28\]
Dividing both sides of equation by \[5\],
\[ \Rightarrow {\left( {x - 1} \right)^2} = \left( {\dfrac{{28}}{5}} \right)\]
Taking square root on both sides,
\[ \Rightarrow \left( {x - 1} \right) = \pm \sqrt {\dfrac{{28}}{5}} \]
\[ \Rightarrow x = 1 \pm \sqrt {\dfrac{{28}}{5}} \]
So, the roots of the equation \[5{x^2} - 10x = 23\] are : \[x = 1 + \sqrt {\dfrac{{28}}{5}} \] and \[x = 1 - \sqrt {\dfrac{{28}}{5}} \] .
Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as $2$. Quadratic equations can be solved by splitting the middle term, factorizing common factors, using the quadratic formula and completing the square method. The given equation can be solved by each and every method listed above.
Complete step-by-step solution:
In the given question, we have to solve the given equation \[5{x^2} - 10x = 23\] using the completing the square method.
So, \[5{x^2} - 10x = 23\]
So, for completing the square, we have to make the given equation resemble the whole square of a binomial identity: ${\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right)$. So, we get,
\[ \Rightarrow {\left( {\sqrt 5 x} \right)^2} - 2\left( {\sqrt 5 x} \right)\left( {\sqrt 5 } \right) = 23\]
Adding the square of the term \[\sqrt 5 \] to both sides of the equation so as to exactly match the above mentioned identity.
\[ \Rightarrow {\left( {\sqrt 5 x} \right)^2} - 2\left( {\sqrt 5 x} \right)\left( {\dfrac{3}{{2\sqrt 5 }}} \right) + {\left( {\sqrt 5 } \right)^2} = 23 + {\left( {\sqrt 5 } \right)^2}\]
So, the three terms on left side of the equation resemble the identity ${\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right)$. Hence, \[\left[ {{{\left( {\sqrt 5 x} \right)}^2} - 2\left( {\sqrt 5 x} \right)\left( {\dfrac{3}{{2\sqrt 5 }}} \right) + {{\left( {\sqrt 5 } \right)}^2}} \right]\] can be condensed as \[{\left[ {\left( {\sqrt 5 x} \right) - \left( {\sqrt 5 } \right)} \right]^2}\]. So, we get,
\[ \Rightarrow {\left( {\sqrt 5 x - \sqrt 5 } \right)^2} = 28\]
Taking out \[5\] from inside the square term,
\[ \Rightarrow 5{\left( {x - 1} \right)^2} = 28\]
Dividing both sides of equation by \[5\],
\[ \Rightarrow {\left( {x - 1} \right)^2} = \left( {\dfrac{{28}}{5}} \right)\]
Taking square root on both sides,
\[ \Rightarrow \left( {x - 1} \right) = \pm \sqrt {\dfrac{{28}}{5}} \]
\[ \Rightarrow x = 1 \pm \sqrt {\dfrac{{28}}{5}} \]
So, the roots of the equation \[5{x^2} - 10x = 23\] are : \[x = 1 + \sqrt {\dfrac{{28}}{5}} \] and \[x = 1 - \sqrt {\dfrac{{28}}{5}} \] .
Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as $2$. Quadratic equations can be solved by splitting the middle term, factorizing common factors, using the quadratic formula and completing the square method. The given equation can be solved by each and every method listed above.
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