Solve \[x(2x - 1) = 0\] ?
Answer
606.6k+ views
Hint: Given equation is in factor form and is equal to 0. Use the concept that a product of two values is equal to zero when either one of them is 0 or both are 0. We equate both factors to 0 one by one and calculate the value of the variable i.e. ‘x’ by shifting the values in the equations.
If \[a \times b = 0\] then either \[a = 0\] or \[b = 0\] or both \[a = 0,b = 0\]
Complete step by step answer:
We are given the equation \[x(2x - 1) = 0\] … (1)
Since the equation is already in factor form then we can directly equate the factors to 0.
We have \[x = 0\] and \[2x - 1 = 0\]
Shift constant value to right hand side of the equation
Then \[x = 0\] and \[2x = 1\]
Divide both sides of the equation by 2
Then \[x = 0\] and \[\dfrac{{2x}}{2} = \dfrac{1}{2}\]
Cancel same factors from numerator and denominator on left hand side of the equation
Then \[x = 0\] and \[x = \dfrac{1}{2}\]
\[\therefore \] Solution of the equation \[x(2x - 1) = 0\] is \[x = 0\] and \[x = \dfrac{1}{2}\] .
Note: Alternate method:
We can open the brackets and make the equation into a quadratic equation.
We can write \[x(2x - 1) = 0\]
\[ \Rightarrow 2{x^2} - x = 0\]
Compare to general quadratic equation \[a{x^2} + bx + c = 0\] we get \[a = 2,b = - 1,c = 0\]
We know when a quadratic equation is equal to 0, we can calculate the roots of the quadratic equation or the value of the variable term in the quadratic equation using a determinant method.
Then roots of the quadratic equation are given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute values of ‘a’, ‘b’ and ‘c’ in the formula
\[ \Rightarrow x = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(2)(0)} }}{{2(2)}}\]
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt 1 }}{4}\]
\[ \Rightarrow x = \dfrac{{1 \pm 1}}{4}\]
Then either \[x = \dfrac{{1 + 1}}{4}\] or \[x = \dfrac{{1 - 1}}{4}\]
Calculate the values in the numerator for both fractions
i.e. either \[x = \dfrac{2}{4}\] or \[x = \dfrac{0}{4}\]
Cancel possible same factors from numerator and denominator of both fractions
i.e. either \[x = \dfrac{1}{2}\] or \[x = 0\]
\[\therefore \] Solution of the equation \[x(2x - 1) = 0\] is \[x = 0\] and \[x = \dfrac{1}{2}\] .
If \[a \times b = 0\] then either \[a = 0\] or \[b = 0\] or both \[a = 0,b = 0\]
Complete step by step answer:
We are given the equation \[x(2x - 1) = 0\] … (1)
Since the equation is already in factor form then we can directly equate the factors to 0.
We have \[x = 0\] and \[2x - 1 = 0\]
Shift constant value to right hand side of the equation
Then \[x = 0\] and \[2x = 1\]
Divide both sides of the equation by 2
Then \[x = 0\] and \[\dfrac{{2x}}{2} = \dfrac{1}{2}\]
Cancel same factors from numerator and denominator on left hand side of the equation
Then \[x = 0\] and \[x = \dfrac{1}{2}\]
\[\therefore \] Solution of the equation \[x(2x - 1) = 0\] is \[x = 0\] and \[x = \dfrac{1}{2}\] .
Note: Alternate method:
We can open the brackets and make the equation into a quadratic equation.
We can write \[x(2x - 1) = 0\]
\[ \Rightarrow 2{x^2} - x = 0\]
Compare to general quadratic equation \[a{x^2} + bx + c = 0\] we get \[a = 2,b = - 1,c = 0\]
We know when a quadratic equation is equal to 0, we can calculate the roots of the quadratic equation or the value of the variable term in the quadratic equation using a determinant method.
Then roots of the quadratic equation are given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute values of ‘a’, ‘b’ and ‘c’ in the formula
\[ \Rightarrow x = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(2)(0)} }}{{2(2)}}\]
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt 1 }}{4}\]
\[ \Rightarrow x = \dfrac{{1 \pm 1}}{4}\]
Then either \[x = \dfrac{{1 + 1}}{4}\] or \[x = \dfrac{{1 - 1}}{4}\]
Calculate the values in the numerator for both fractions
i.e. either \[x = \dfrac{2}{4}\] or \[x = \dfrac{0}{4}\]
Cancel possible same factors from numerator and denominator of both fractions
i.e. either \[x = \dfrac{1}{2}\] or \[x = 0\]
\[\therefore \] Solution of the equation \[x(2x - 1) = 0\] is \[x = 0\] and \[x = \dfrac{1}{2}\] .
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