Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) \[2{x^2} - 7x + 3 = 0\] (ii) \[2{x^2} + x - 4 = 0\]
(iii) \[4{x^2} + 4\sqrt 3 x + 3 = 0\] (iv) \[2{x^2} + x + 4 = 0\]

Answer 1

Hint: Here, we need to find the roots of the given quadratic equations by completing the square. First, we will divide the equation by a number such that the coefficient of \[{x^2}\] is 1. Then, we will add and subtract the square of the half of the coefficient of \[x\]. Then, using the algebraic identity for the square of the sum or difference of two numbers, we will complete the square. Then, we will simplify the expression, apply the algebraic identity for the product of the sum and difference of two numbers, and equate both parentheses to zero to get the two roots of the quadratic equation.

Formula Used: We will use the following formulas to solve the question:
The square of the difference of two numbers is given by the algebraic identity \[{a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}\].
The square of the sum of two numbers is given by the algebraic identity \[{a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\].
The product of the sum and difference of two numbers is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

Complete step-by-step answer:
(i)
The given quadratic equation is \[2{x^2} - 7x + 3 = 0\].
First, we will divide the equation by 2 such that the coefficient of \[{x^2}\] is 1.
Dividing both sides by 2, we get
\[ \Rightarrow \dfrac{{2{x^2} - 7x + 3}}{2} = \dfrac{0}{2} \\
   \Rightarrow {x^2} - \dfrac{7}{2}x + \dfrac{3}{2} = 0 \\\]
Now, to complete the square, we will add and subtract the square of the half of the coefficient of \[x\].
We can observe that the coefficient of \[x\] is \[\dfrac{7}{2}\].
The half of \[\dfrac{7}{2}\] is \[\dfrac{{\dfrac{7}{2}}}{2} = \dfrac{7}{4}\].
Thus, we will add and subtract \[{\left( {\dfrac{7}{4}} \right)^2}\] from the left hand side of the equation.
Therefore, we get
\[ \Rightarrow {x^2} - \dfrac{7}{2}x + \dfrac{3}{2} + {\left( {\dfrac{7}{4}} \right)^2} - {\left( {\dfrac{7}{4}} \right)^2} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{7}{4}} \right)^2} - 2\left( x \right)\left( {\dfrac{7}{4}} \right) + \dfrac{3}{2} - {\left( {\dfrac{7}{4}} \right)^2} = 0\]
Now, we know that the square of the difference of two numbers is given by the algebraic identity \[{a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}\].
Substituting \[a = x\] and \[b = \dfrac{7}{4}\] in the identity, we get
\[{\left( x \right)^2} + {\left( {\dfrac{7}{4}} \right)^2} - 2\left( x \right)\left( {\dfrac{7}{4}} \right)x = {\left( {x - \dfrac{7}{4}} \right)^2}\]
Substituting \[{\left( x \right)^2} + {\left( {\dfrac{7}{4}} \right)^2} - 2\left( x \right)\left( {\dfrac{7}{4}} \right)x = {\left( {x - \dfrac{7}{4}} \right)^2}\] in the equation \[{\left( x \right)^2} + {\left( {\dfrac{7}{4}} \right)^2} - 2\left( x \right)\left( {\dfrac{7}{4}} \right)x + \dfrac{3}{2} - {\left( {\dfrac{7}{4}} \right)^2} = 0\], we get
\[ \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} + \dfrac{3}{2} - {\left( {\dfrac{7}{4}} \right)^2} = 0\]
Simplifying the expression, we get
\[\Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} + \dfrac{3}{2} - \dfrac{{49}}{{16}} = 0 \\
   \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} + \dfrac{{24}}{{16}} - \dfrac{{49}}{{16}} = 0 \\
   \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} - \dfrac{{25}}{{16}} = 0 \\\]
Rewriting the expression \[\dfrac{{25}}{{16}}\], we get
\[ \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} - {\left( {\dfrac{5}{4}} \right)^2} = 0\]
Now, the product of the sum and difference of two numbers is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].
Therefore, applying the algebraic identity in the equation, we get
\[ \Rightarrow \left( {x - \dfrac{7}{4} + \dfrac{5}{4}} \right)\left( {x - \dfrac{7}{4} - \dfrac{5}{4}} \right) = 0\]
Thus, either \[x - \dfrac{7}{4} + \dfrac{5}{4} = 0\] or \[x - \dfrac{7}{4} - \dfrac{5}{4} = 0\].
Simplifying the expressions, we get
\[ \Rightarrow x - \dfrac{2}{4} = 0\] or \[x - \dfrac{{12}}{4} = 0\]
\[ \Rightarrow x - \dfrac{1}{2} = 0\] or \[x - 3 = 0\]
\[ \Rightarrow x = \dfrac{1}{2}\] or \[x = 3\]
\[\therefore \] The roots of the quadratic equation \[2{x^2} - 7x + 3 = 0\] are \[\dfrac{1}{2}\] and 3.
(ii)
The given quadratic equation is \[2{x^2} + x - 4 = 0\].
First, we will divide the equation by 2 such that the coefficient of \[{x^2}\] is 1.
Dividing both sides by 2, we get
\[\Rightarrow \dfrac{{2{x^2} + x - 4}}{2} = \dfrac{0}{2} \\
   \Rightarrow {x^2} + \dfrac{1}{2}x - \dfrac{4}{2} = 0 \\
   \Rightarrow {x^2} + \dfrac{1}{2}x - 2 = 0 \\\]
Now, to complete the square, we will add and subtract the square of the half of the coefficient of \[x\].
We can observe that the coefficient of \[x\] is \[\dfrac{1}{2}\].
The half of \[\dfrac{1}{2}\] is \[\dfrac{{\dfrac{1}{2}}}{2} = \dfrac{1}{4}\].
Thus, we will add and subtract \[{\left( {\dfrac{1}{4}} \right)^2}\] from the left hand side of the equation.
Therefore, we get
\[ \Rightarrow {x^2} + \dfrac{1}{2}x - 2 + {\left( {\dfrac{1}{4}} \right)^2} - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right) - 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Now, we know that the square of the sum of two numbers is given by the algebraic identity \[{a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\].
Substituting \[a = x\] and \[b = \dfrac{1}{4}\] in the identity, we get
\[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right)x = {\left( {x + \dfrac{1}{4}} \right)^2}\]
Substituting \[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right)x = {\left( {x + \dfrac{1}{4}} \right)^2}\] in the equation \[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right) - 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\], we get
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Simplifying the expression, we get
\[\Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - 2 - \dfrac{1}{{16}} = 0 \\
   \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - \dfrac{{32}}{{16}} - \dfrac{1}{{16}} = 0 \\
   \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - \dfrac{{33}}{{16}} = 0 \\\]
Rewriting the expression \[\dfrac{{33}}{{16}}\], we get
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} - {\left( {\dfrac{{\sqrt {33} }}{4}} \right)^2} = 0\]
Now, the product of the sum and difference of two numbers is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].
Therefore, applying the algebraic identity in the equation, we get
\[ \Rightarrow \left( {x + \dfrac{1}{4} + \dfrac{{\sqrt {33} }}{4}} \right)\left( {x + \dfrac{1}{4} - \dfrac{{\sqrt {33} }}{4}} \right) = 0\]
Thus, either \[x + \dfrac{1}{4} + \dfrac{{\sqrt {33} }}{4} = 0\] or \[x + \dfrac{1}{4} - \dfrac{{\sqrt {33} }}{4} = 0\].
Simplifying the expressions, we get
\[ \Rightarrow x = - \dfrac{1}{4} - \dfrac{{\sqrt {33} }}{4}\] or \[x = - \dfrac{1}{4} + \dfrac{{\sqrt {33} }}{4}\]
\[ \Rightarrow x = \dfrac{{ - 1 - \sqrt {33} }}{4}\] or \[x = \dfrac{{ - 1 + \sqrt {33} }}{4}\]
\[\therefore \] The roots of the quadratic equation
\[2{x^2} + x - 4 = 0\] are \[\dfrac{{ - 1 - \sqrt {33} }}{4}\] and \[\dfrac{{ - 1 + \sqrt {33} }}{4}\].
(iii)
The given quadratic equation is \[4{x^2} + 4\sqrt 3 x + 3 = 0\].
First, we will divide the equation by 4 such that the coefficient of \[{x^2}\] is 1.
Dividing both sides by 4, we get
\[ \Rightarrow \dfrac{{4{x^2} + 4\sqrt 3 x + 3}}{4} = \dfrac{0}{4} \\
   \Rightarrow {x^2} + \sqrt 3 x + \dfrac{3}{4} = 0 \\\]
Now, to complete the square, we will add and subtract the square of the half of the coefficient of \[x\].
We can observe that the coefficient of \[x\] is \[\sqrt 3 \].
The half of \[\sqrt 3 \] is \[\dfrac{{\sqrt 3 }}{2}\].
Thus, we will add and subtract \[{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}\] from the left hand side of the equation.
Therefore, we get
\[ \Rightarrow {x^2} + \sqrt 3 x + \dfrac{3}{4} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + 2\left( x \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \dfrac{3}{4} - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = 0\]
Now, we know that the square of the sum of two numbers is given by the algebraic identity \[{a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\].
Substituting \[a = x\] and \[b = \dfrac{{\sqrt 3 }}{2}\] in the identity, we get
\[{\left( x \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + 2\left( x \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)x = {\left( {x + \dfrac{{\sqrt 3 }}{2}} \right)^2}\]
Substituting \[{\left( x \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + 2\left( x \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)x = {\left( {x + \dfrac{{\sqrt 3 }}{2}} \right)^2}\] in the equation \[{\left( x \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + 2\left( x \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \dfrac{3}{4} - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = 0\], we get
\[ \Rightarrow {\left( {x + \dfrac{{\sqrt 3 }}{2}} \right)^2} + \dfrac{3}{4} - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = 0\]
Simplifying the expression, we get
\[\Rightarrow {\left( {x + \dfrac{{\sqrt 3 }}{2}} \right)^2} + \dfrac{3}{4} - \dfrac{3}{4} = 0 \\
   \Rightarrow {\left( {x + \dfrac{{\sqrt 3 }}{2}} \right)^2} = 0 \\
   \Rightarrow \left( {x + \dfrac{{\sqrt 3 }}{2}} \right)\left( {x + \dfrac{{\sqrt 3 }}{2}} \right) = 0 \\\]
Thus, the roots of the quadratic equation are equal.
Therefore, we get
 \[ \Rightarrow x + \dfrac{{\sqrt 3 }}{2} = 0\]
Simplifying the expression, we get
\[ \Rightarrow x = - \dfrac{{\sqrt 3 }}{2}\]
\[\therefore \] The roots of the quadratic equation
\[4{x^2} + 4\sqrt 3 x + 3 = 0\] are \[ - \dfrac{{\sqrt 3 }}{2}\] and \[ - \dfrac{{\sqrt 3 }}{2}\].
(iv)
The given quadratic equation is \[2{x^2} + x + 4 = 0\].
First, we will divide the equation by 2 such that the coefficient of \[{x^2}\] is 1.
Dividing both sides by 2, we get
\[\Rightarrow \dfrac{{2{x^2} + x + 4}}{2} = \dfrac{0}{2} \\
   \Rightarrow {x^2} + \dfrac{1}{2}x + \dfrac{4}{2} = 0 \\
   \Rightarrow {x^2} + \dfrac{1}{2}x + 2 = 0 \\\]
Now, to complete the square, we will add and subtract the square of the half of the coefficient of \[x\].
We can observe that the coefficient of \[x\] is
\[\dfrac{1}{2}\].
The half of \[\dfrac{1}{2}\] is \[\dfrac{{\dfrac{1}{2}}}{2} = \dfrac{1}{4}\].
Thus, we will add and subtract \[{\left( {\dfrac{1}{4}} \right)^2}\] from the left hand side of the equation.
Therefore, we get
\[ \Rightarrow {x^2} + \dfrac{1}{2}x + 2 + {\left( {\dfrac{1}{4}} \right)^2} - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right) + 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Now, we know that the square of the sum of two numbers is given by the algebraic identity \[{a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}\].
Substituting \[a = x\] and \[b = \dfrac{1}{4}\] in the identity, we get
\[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right)x = {\left( {x + \dfrac{1}{4}} \right)^2}\]
Substituting \[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right)x = {\left( {x + \dfrac{1}{4}} \right)^2}\] in the equation \[{\left( x \right)^2} + {\left( {\dfrac{1}{4}} \right)^2} + 2\left( x \right)\left( {\dfrac{1}{4}} \right) + 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\], we get
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} + 2 - {\left( {\dfrac{1}{4}} \right)^2} = 0\]
Simplifying the expression, we get
\[\Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} + 2 - \dfrac{1}{{16}} = 0 \\
   \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} + \dfrac{{32}}{{16}} - \dfrac{1}{{16}} = 0 \\
   \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} + \dfrac{{31}}{{16}} = 0 \\\]
Rewriting the expression, we get
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} = - \dfrac{{31}}{{16}}\]
We know that the square of any real number is greater than or equal to 0.
Therefore, \[{\left( {x + \dfrac{1}{4}} \right)^2}\] cannot be equal to \[ - \dfrac{{31}}{{16}}\].
Since there is no real value of \[x\] for which \[{\left( {x + \dfrac{1}{4}} \right)^2} = - \dfrac{{31}}{{16}}\], the roots of the quadratic equation \[2{x^2} + x + 4 = 0\] do not exist.

Note: We can also calculate the discriminant before starting the completing the square method to check whether the roots of the quadratic equation exist or not. The discriminant of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by \[D = {b^2} - 4ac\]. If the discriminant is negative, then no real roots of the quadratic equation exist.