#### Question: Find the roots of the following quadratic equations, if they exist, by the method of completing squares:

i. 2x^{2} - 7x + 3 = 0

ii. 2x^{2} + x – 4 = 0

iii. 4x^{2}+ 4\\sqrt3)x + 3 = 0

iv. 2x^{2} + x + 4 = 0

#### Video Explanation for i. 2x^{2} - 7x + 3 = 0

#### Explanatory Answer

##### i. 2x^{2} - 7x + 3 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x^{2} is 1.

x^{2} - \\frac{7}{2})x + \\frac{3}{2}) = 0

Step 2: Divide the coefficient of x by 2. i.e., \\frac{–7}{2}) by 2 and complete the square as shown below.

(x - \\frac {7} {4}{)}^{2}) – \(\frac{7}{4}{)}^{2}) + \\frac{3}{2}) = 0

(x - \\frac {7} {4}{)}^{2}) - \\frac{49}{16}) + \\frac{3}{2}) = 0

(x - \\frac {7} {4}{)}^{2}) - \\frac{49 - 24}{16}) = 0

(x - \\frac {7} {4}{)}^{2}) = \\frac{25}{16})

x - \\frac{7}{4}) = ±\\frac{5}{4})

x = \\frac{5}{4}) + \\frac{7}{4}) and x = -\\frac{5}{4}) + \\frac{7}{4})

x = 3 and x = \\frac{1}{2})

##### i. 2x^{2} - 7x + 3 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x^{2} is a perfect square.

4x^{2} – 14x + 6 = 0

Step 2: We have to expres 4x^{2} – 14x + 6 = 0 as a square of the form (p - q)^{2} ± something = 0.

Let us find out what p and q are. 4x^{2} is p^{2}

∴ p = 2x.

So, the square will be (2x - q)^{2}

Only step left to complete the square is find q. Let us find that.

2pq = 14x

p = 2x. So, 4xq = 14x. Or q = \\frac{7}{2})

So, the square is completed as shown below

(2x - \\frac {7} {2}{)}^{2}) – \(\frac{7}{2}{)}^{2}) + 6 = 0

(2x - \\frac {7} {2}{)}^{2}) – \\frac{49}{4}) + 6 = 0

(2x - \\frac {7} {2}{)}^{2}) - \\frac{49 - 24}{4}) = 0

(2x - \\frac {7} {2}{)}^{2}) – \\frac{25}{4}) = 0

(2x - \\frac {7} {2}{)}^{2}) = \\frac{25}{4})

(2x – \\frac{7}{2})) = ±\\frac{5}{2})

2x = \\frac{7}{2}) + \\frac{5}{2}) and 2x = \\frac{7}{2}) – \\frac{5}{2})

2x = 6 and 2x = 1

x = 3 and x = \\frac{1}{2})

#### Video Explanation for ii, iii, and iv

##### ii. 2x^{2} + x – 4 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x^{2} is 1.

x^{2} + \\frac{x}{2}) – 2 = 0

Step 2: Divide the coefficient of x by 2. i.e., \\frac{1}{2}) by 2 and complete the square as shown below.

(x - \\frac {1} {4}{)}^{2}) – \(\frac{1}{4}{)}^{2}) – 2 = 0

(x - \\frac {1} {4}{)}^{2}) – \\frac{1}{16}) - 2 = 0

(x - \\frac {1} {4}{)}^{2}) – \\frac{33}{16}) = 0

(x - \\frac {1} {4}{)}^{2}) = \\frac{33}{16})

Or x + \\frac{1}{4}) = ± \\frac{\sqrt{33}}{4})

x = \(\frac{-1 + \sqrt{33}}{4})) or x = \(\frac{-1 - \sqrt{33}}{4}))

##### ii. 2x^{2} + x – 4 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x^{2} is a perfect square.

4x^{2} + 2x – 8 = 0

Step 2: We have to expres 4x^{2} + 2x – 8 = 0 as a square of the form (p + q)^{2} ± something = 0.

Let us find out what p and q are. 4x^{2} is p^{2}

∴ p = 2x.

So, the square will be (2x + q)^{2}

Only step left to complete the square is find q. Let us find that.

2pq = 2x

p = 2x. So, 4xq = 2x. Or q = \\frac{1}{2})

So, the square is completed as shown below

(2x + \\frac {1} {2}{)}^{2}) – \(\frac{1}{2}{)}^{2}) – 8 = 0

(2x + \\frac {1} {2}{)}^{2}) – \\frac{1}{4}) - 8 = 0

(2x + \\frac {1} {2}{)}^{2}) – \\frac{33}{4}) = 0

(2x + \\frac {1} {2}{)}^{2}) = \\frac{33}{4})

\\frac{2x+1}{2}) = ±\\frac{\sqrt33}{2})

x = ±\\frac{{\sqrt33} - 1}{4}) or x = \\frac{{-\sqrt33} - 1}{4})

##### iii. 4x^{2} + 4√3 x + 3 = 0 | Method 1

Step 1: Divide the entire equation by 4 so that the coefficient of x^{2} is 1.

x^{2} + \\sqrt{3})x + \\frac{3}{4}) = 0

Step 2: Divide the coefficient of x by 2. i.e., \\sqrt3) by 2 and complete the square as shown below.

(x + \\frac {\sqrt3} {2}{)}^{2}) - \(\frac{\sqrt3}{2}{)}^{2}) + \\frac{3}{4}) = 0

(x + \\frac {\sqrt3} {2}{)}^{2}) -\\frac{3}{4}) + \\frac{3}{4}) = 0

(x + \\frac {\sqrt3} {2}{)}^{2}) = 0

x = - \\frac{\sqrt3}{2}) and -\\frac{\sqrt3}{2})

##### iii. 4x^{2} + 4√3 x + 3 = 0 | Method 2

Step 1: The coefficient of x^{2} is a perfect square. So, there is nothing to be multiplied.

Step 2: We have to expres 4x^{2} + 4\\sqrt{3})x + 3 = 0 as a square of the form (p + q)^{2} ± something = 0.

Let us find out what p and q are. 4x^{2} is p^{2}

∴ p = 2x.

So, the square will be (2x + q)^{2}

Only step left to complete the square is find q. Let us find that.

2pq = 4\\sqrt{3})x

p = 2x. So, 4xq = 4\\sqrt{3})x. Or q = \\sqrt{3})

So, the square is completed as shown below

(2x + \\sqrt3))^{2} - (\\sqrt3))^{2} + 3 = 0

(2x + \\sqrt3))^{2} - 3 + 3 = 0

(2x + \\sqrt3))^{2} = 0

2x + \\sqrt3) = 0

x = -\\frac{\sqrt3}{2}) and - \\frac{\sqrt3}{2})

##### iv. 2x^{2}+ x + 4 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x^{2} is 1.

x^{2} + \\frac{x}{2}) + 2 = 0

Step 2: Divide the coefficient of x by 2. i.e., \\frac{1}{2}) by 2 and complete the square as shown below.

(x + \\frac {1} {4}{)}^{2}) - \(\frac{1}{4}{)}^{2}) + 2 = 0

(x + \\frac {1} {4}{)}^{2}) - \\frac{1}{16}) + 2 = 0

(x + \\frac {1} {4}{)}^{2}) + \\frac{32 – 1}{16}) = 0

(x + \\frac {1} {4}{)}^{2}) = -\\frac{31}{16}) < 0

(x + \\frac {1} {4}{)}^{2}) cannot be negative for any real value of x because square of real numbers is always non - negative.

∴ the given equation has no real roots.

##### iv. 2x^{2}+ x + 4 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x^{2 } is a perfect square.

4x^{2 } + 2x + 8 = 0

Step 2: We have to expres 4x^{2 } + 2x + 8 = 0 as a square of the form (p + q)^{2} ± something = 0.

Let us find out what p and q are. 4x^{2} is p^{2}

∴ p = 2x.

So, the square will be (2x + q)^{2}

Only step left to complete the square is find q. Let us find that.

2pq = 2x

p = 2x. So, 4xq = 2x. Or q = \\frac{1}{2})

So, the square is completed as shown below

(2x + \\frac {1} {2}{)}^{2}) - \(\frac{1}{2}{)}^{2}) + 8 = 0

(2x + \\frac {1} {2}{)}^{2}) - \\frac{1}{4}) + 8 = 0

(2x + \\frac {1} {2}{)}^{2}) + \\frac{32-1}{4}) = 0

(2x + \\frac {1} {2}{)}^{2}) = -\\frac{31}{4}) < 0

(2x + \\frac {1} {2}{)}^{2}) cannot be negative for any real value of x because square of real numbers is always non - negative.

∴ the given equation has no real roots.