
How do you solve $ 2{x^2} - 7x + 3 = 0 $ by completing the square?
Answer
497.1k+ views
Hint: In this question we need to find the roots of the quadratic equation $ 2{x^2} - 7x + 3 = 0 $ by completing the square method. In completing square method you have to use these basic algebraic identities $ {(a + b)^2} = {a^2} + 2ab + {b^2},{(a - b)^2} = {a^2} - 2ab + {b^2} $ , you need to convert given quadratic polynomial in the form of $ {(a + b)^2} $ or $ {(a - b)^2} $ and after which you will get the root of the given equation.
Complete step-by-step answer:
Let us try to solve this question, the quadratic equation for which we have to find the roots is $ 2{x^2} - 7x + 3 = 0 $ .
Here we have to find the root of the quadratic equation by completing the square method.
Since in the given quadratic equation $ 2{x^2} - 7x + 3 = 0 $ , the coefficient of $ x $ is negative. So we will try to convert this quadratic equation in the form of algebraic identity $ {(a - b)^2} $ .
$ 2{x^2} - 7x + 3 = 0 $
$ $ $ {x^2} - \dfrac{{7x}}{2} + \dfrac{3}{2} = 0 $
Is same as $ {(x)^2} - 2 \times x \times \dfrac{7}{4} + {\left( {\dfrac{7}{4}} \right)^2} - {\left( {\dfrac{7}{4}} \right)^2} + \dfrac{3}{2} = 0 $ by using identity $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ . We will get
$ {\left( {x - \dfrac{7}{4}} \right)^2} - \dfrac{{49}}{{16}} + \dfrac{3}{2} = 0 $
$ {\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{25}}{{16}} $
Taking square root on both side we will get,
$ \left( {x - \dfrac{7}{4}} \right) = \pm \dfrac{5}{4} $
Now adding $ \dfrac{7}{4} $ in both L.H.S and R.H.S of above equation we will get
\[
\left( {x - \dfrac{7}{4}} \right) + \dfrac{7}{4} = \pm \dfrac{5}{4} + \dfrac{7}{4} \\
x = \pm \dfrac{5}{4} + \dfrac{7}{4} \\
\]
Hence the roots of the given quadratic equation $ 2{x^2} - 7x + 3 = 0 $ are $ x = 3 $ and $ x = \dfrac{1}{2} $ .
Both the roots of the given quadratic equation are real and not equal.
So, the correct answer is “ $ x = 3 $ and $ x = \dfrac{1}{2} $ ”.
Note: We can find roots of quadratic equations using factorization method also. But in this question are specifically asked to solve using completing the square method. A common mistake made by students while solving for roots of quadratic equation by completing the square method is when they are taking root on both sides of the equation, the square root of a number can either be a positive or negative number. So, please be careful while taking square roots.
Complete step-by-step answer:
Let us try to solve this question, the quadratic equation for which we have to find the roots is $ 2{x^2} - 7x + 3 = 0 $ .
Here we have to find the root of the quadratic equation by completing the square method.
Since in the given quadratic equation $ 2{x^2} - 7x + 3 = 0 $ , the coefficient of $ x $ is negative. So we will try to convert this quadratic equation in the form of algebraic identity $ {(a - b)^2} $ .
$ 2{x^2} - 7x + 3 = 0 $
$ $ $ {x^2} - \dfrac{{7x}}{2} + \dfrac{3}{2} = 0 $
Is same as $ {(x)^2} - 2 \times x \times \dfrac{7}{4} + {\left( {\dfrac{7}{4}} \right)^2} - {\left( {\dfrac{7}{4}} \right)^2} + \dfrac{3}{2} = 0 $ by using identity $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ . We will get
$ {\left( {x - \dfrac{7}{4}} \right)^2} - \dfrac{{49}}{{16}} + \dfrac{3}{2} = 0 $
$ {\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{25}}{{16}} $
Taking square root on both side we will get,
$ \left( {x - \dfrac{7}{4}} \right) = \pm \dfrac{5}{4} $
Now adding $ \dfrac{7}{4} $ in both L.H.S and R.H.S of above equation we will get
\[
\left( {x - \dfrac{7}{4}} \right) + \dfrac{7}{4} = \pm \dfrac{5}{4} + \dfrac{7}{4} \\
x = \pm \dfrac{5}{4} + \dfrac{7}{4} \\
\]
Hence the roots of the given quadratic equation $ 2{x^2} - 7x + 3 = 0 $ are $ x = 3 $ and $ x = \dfrac{1}{2} $ .
Both the roots of the given quadratic equation are real and not equal.
So, the correct answer is “ $ x = 3 $ and $ x = \dfrac{1}{2} $ ”.
Note: We can find roots of quadratic equations using factorization method also. But in this question are specifically asked to solve using completing the square method. A common mistake made by students while solving for roots of quadratic equation by completing the square method is when they are taking root on both sides of the equation, the square root of a number can either be a positive or negative number. So, please be careful while taking square roots.
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