Quadratic Equations | Completing Squares

Explanatory Answer

i. 2x² - 7x + 3 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x² is 1.
x² - \\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. 2x² - 7x + 3 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x² is a perfect square.
4x² – 14x + 6 = 0

Step 2: We have to expres 4x² – 14x + 6 = 0 as a square of the form (p - q)² ± something = 0.
Let us find out what p and q are. 4x² is p²
∴ 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})

ii. 2x² + x – 4 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x² is 1.
x² + \\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. 2x² + x – 4 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x² is a perfect square.
4x² + 2x – 8 = 0

Step 2: We have to expres 4x² + 2x – 8 = 0 as a square of the form (p + q)² ± something = 0.
Let us find out what p and q are. 4x² is p²
∴ 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. 4x² + 4√3 x + 3 = 0 | Method 1

Step 1: Divide the entire equation by 4 so that the coefficient of x² is 1.
x² + \\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. 4x² + 4√3 x + 3 = 0 | Method 2

Step 1: The coefficient of x² is a perfect square. So, there is nothing to be multiplied.

Step 2: We have to expres 4x² + 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. 4x² is p²
∴ 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. 2x²+ x + 4 = 0 | Method 1

Step 1: Divide the entire equation by 2 so that the coefficient of x² is 1.
x² + \\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. 2x²+ x + 4 = 0 | Method 2

Step 1: Multiply the entire equation by 2 so that the coefficient of x² is a perfect square.
4x² + 2x + 8 = 0

Step 2: We have to expres 4x² + 2x + 8 = 0 as a square of the form (p + q)² ± something = 0.
Let us find out what p and q are. 4x² is p²
∴ 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.