Question

How do you solve using the complete square root method $2{x^2} - 3x + 1 = 0$

Answer 1

Hint: This problem comes under solving quadratic equations in algebra. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. There are methods to solve quadratic equations; they are factorisation method, formula method and complete square root method. Here we asked to solve by the complete square root method. There are steps to solve by the complete square root method. By using condition and complete step by step explanation we solve this.

Complete step-by-step solution:
Step 1: Write the quadratic equation in the general form \[a{x^2} + bx + c = 0\]
Step 2: Divide the both sides of the equation by the coefficient of \[{x^2}\]if it is not 1.
Step 3: Shift the constant term to the right hand side.
Step 4: Add the square of one-half of the co-efficient to \[x\]to both sides.
Step 5: Write the left hand side as a square and simplify the right hand side
Step 6: Take the square root on both sides and solve \[x\]
Let us consider the equation $2{x^2} - 3x + 1 = 0$
Step 1: $\dfrac{{2{x^2}}}{2} - \dfrac{{3x}}{2} + \dfrac{1}{2} = 0$
Step 2: \[{x^2} - \dfrac{{3x}}{2} = - \dfrac{1}{2}\]\[x = 1,\dfrac{1}{2}\]
Step 3: \[{x^2} - \dfrac{{3x}}{2} + {\left( {\dfrac{3}{4}} \right)^2} = - \dfrac{1}{2} + {\left( {\dfrac{3}{4}} \right)^2}\]
Now left hand side of the equation looks like\[{\left( {a - b} \right)^2} = {a^2} + 2ab + {b^2}\], so convert\[ \Rightarrow {x^2} - \dfrac{{3x}}{2} + {\left( {\dfrac{3}{4}} \right)^2} = {\left( {x - \dfrac{3}{4}} \right)^2}\]
Step 4: \[{\left( {x - \dfrac{3}{4}} \right)^2} = - \dfrac{1}{2} + \dfrac{9}{{16}}\]
Taking Square root on both sides, we get
Step 5: \[\sqrt {{{\left( {x - \dfrac{3}{4}} \right)}^2}} = \sqrt { - \dfrac{8}{{16}} + \dfrac{9}{{16}}} \]
\[ \Rightarrow \left( {x - \dfrac{3}{4}} \right) = \sqrt {\dfrac{1}{{16}}} \]
\[ \Rightarrow x - \dfrac{3}{4} = \pm \dfrac{1}{4}\]
Now we separate \[x\]and solve
\[ \Rightarrow x = \pm \dfrac{1}{4} + \dfrac{3}{4}\]
\[ \Rightarrow x = + \dfrac{1}{4} + \dfrac{3}{4}, - \dfrac{1}{4} + \dfrac{3}{4}\]
\[ \Rightarrow x = \dfrac{4}{4},\dfrac{2}{4}\]
\[ \Rightarrow x = 1,\dfrac{1}{2}\]

Therefore the answers are \[x = 1,\dfrac{1}{2}\]

Note: Here we find the roots of the given equation by complete square root method. We can also determine the nature of roots by delta. \[\Delta = {b^2} - 4ac\],
If $\Delta = 0,$ the roots are real and equal
If \[\Delta > 0,\] the roots are real and unequal
If \[\Delta < 0,\] the roots are imaginary
By this we can find the nature of roots. The \[\Delta = {b^2} - 4ac\] value is known as discriminant.