2x² + x – 4 = 0 Diving both sides by 2 x²+ 1/2 x - 2 = 0 ⇒ x² + 1/2 x = 2 Adding [1/2(1/2)]² on both the sides, we get x² + 1/2 x + (1/4)² = 2 + (1/4)² [As x = (-b±√(b²-4ac))/2a] ⇒ (x + 1/4)² = 2 + 1/16 ⇒ (x + 1/4)² = 32 + 1/16 ⇒ (x + 1/4)² = 33/16 ⇒ x + 1/4 = ±√33/4 Either x + 1/4 = √33/4 or x + 1/Read more
2x² + x – 4 = 0
Diving both sides by 2
x²+ 1/2 x – 2 = 0
⇒ x² + 1/2 x = 2
Adding [1/2(1/2)]² on both the sides, we get
x² + 1/2 x + (1/4)² = 2 + (1/4)² [As x = (-b±√(b²-4ac))/2a]
⇒ (x + 1/4)² = 2 + 1/16 ⇒ (x + 1/4)² = 32 + 1/16
⇒ (x + 1/4)² = 33/16 ⇒ x + 1/4 = ±√33/4
Either x + 1/4 = √33/4 or x + 1/4 = – √33/4
⇒ x = √33/4 – 1/4 or x = – √33/4 – 1/4
⇒ x = √33-1/4 or x = -√33-1/4
Hence, the roots of the quadratic equation are √33-1/4 and -√33-1/4.
Let the Rehman's current age = x years Therefore, 3 years ago, age = x -3 years and, after 5 years, age = x + 5 years according to questions, 1/x -3 + 1/x + 5 = 1/3 ⇒ (x +5) + (x - 3)/(x - 3)(x +5) = 1/3 ⇒ 2x + 2/x² + 2x - 15 = 1/3 ⇒ x² + 2x - 15 = 6x + 6 ⇒ x² - 4x - 21 = 0 For the quadratic equatioRead more
Let the Rehman’s current age = x years
Therefore, 3 years ago, age = x -3 years
and, after 5 years, age = x + 5 years
according to questions,
1/x -3 + 1/x + 5 = 1/3
⇒ (x +5) + (x – 3)/(x – 3)(x +5) = 1/3
⇒ 2x + 2/x² + 2x – 15 = 1/3
⇒ x² + 2x – 15 = 6x + 6
⇒ x² – 4x – 21 = 0
For the quadratic equation x² – 4x – 21 = 0, we have a = 1, b = – 4 and c = – 21.
Therefore, b² – 4ac = (-4)² – 4 × 1 × (-21) = 16 + 84 = 100 > 0
Hence, x = 4±√100/2 = 4±10/2 [As x = (-b±√(b²-4ac))/2a]
Either x = 4+10/2 = 14/2 = 7 or x = 4 -10/2 = – 6/2 = – 3
Age of person can’t be negative, so Rehman’s age = 7 years.
2x² + x + 4 = 0 dividing both sides by 2 x² + 1/2 x = - 2 Adding [1/2(1/2)]² on both the sides, we get x² + 1/2 x + (1/4)² = - 2 + (1/4)² [ As x = (-b±√b²- 4ac)/2a] ⇒ (x+1/4)² = - 2 + 1/16 ⇒ (x + 1/4)² = (-32 + 1)/16 ⇒ (x + 1/4)² = - 31/16 < 0 we know that the square of any real number can not beRead more
2x² + x + 4 = 0
dividing both sides by 2
x² + 1/2 x = – 2
Adding [1/2(1/2)]² on both the sides, we get
x² + 1/2 x + (1/4)² = – 2 + (1/4)² [ As x = (-b±√b²- 4ac)/2a]
⇒ (x+1/4)² = – 2 + 1/16 ⇒ (x + 1/4)² = (-32 + 1)/16
⇒ (x + 1/4)² = – 31/16 < 0
we know that the square of any real number can not be negative or (x + 1/4)² can not be negative. Hence, this quadratic equation has no real roots.
4x² + 4√3x + 3 = 0 For the quadratic equation 4x² + 4√3x + 3 = 0, we have a = 2, b = 1, and c = 4. Therefore, b² - 4ac = (4√3)² - 4 × 4 × 3 = 48 - 48 = 0 Hence, x = (-4√3±√0)/8 = -4√3/8 = - √3/2 [As, x = (-b±√b²-4ac)/2a] or x = -√3/2 Hence, the roots of quadratic equation are - √3/2 and - √3/2.
4x² + 4√3x + 3 = 0
For the quadratic equation 4x² + 4√3x + 3 = 0, we have a = 2, b = 1, and c = 4.
Therefore, b² – 4ac = (4√3)² – 4 × 4 × 3 = 48 – 48 = 0
Hence, x = (-4√3±√0)/8 = -4√3/8 = – √3/2 [As, x = (-b±√b²-4ac)/2a]
or x = -√3/2
Hence, the roots of quadratic equation are – √3/2 and – √3/2.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square: 2x² + x – 4 = 0
2x² + x – 4 = 0 Diving both sides by 2 x²+ 1/2 x - 2 = 0 ⇒ x² + 1/2 x = 2 Adding [1/2(1/2)]² on both the sides, we get x² + 1/2 x + (1/4)² = 2 + (1/4)² [As x = (-b±√(b²-4ac))/2a] ⇒ (x + 1/4)² = 2 + 1/16 ⇒ (x + 1/4)² = 32 + 1/16 ⇒ (x + 1/4)² = 33/16 ⇒ x + 1/4 = ±√33/4 Either x + 1/4 = √33/4 or x + 1/Read more
2x² + x – 4 = 0
See lessDiving both sides by 2
x²+ 1/2 x – 2 = 0
⇒ x² + 1/2 x = 2
Adding [1/2(1/2)]² on both the sides, we get
x² + 1/2 x + (1/4)² = 2 + (1/4)² [As x = (-b±√(b²-4ac))/2a]
⇒ (x + 1/4)² = 2 + 1/16 ⇒ (x + 1/4)² = 32 + 1/16
⇒ (x + 1/4)² = 33/16 ⇒ x + 1/4 = ±√33/4
Either x + 1/4 = √33/4 or x + 1/4 = – √33/4
⇒ x = √33/4 – 1/4 or x = – √33/4 – 1/4
⇒ x = √33-1/4 or x = -√33-1/4
Hence, the roots of the quadratic equation are √33-1/4 and -√33-1/4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.
Let the Rehman's current age = x years Therefore, 3 years ago, age = x -3 years and, after 5 years, age = x + 5 years according to questions, 1/x -3 + 1/x + 5 = 1/3 ⇒ (x +5) + (x - 3)/(x - 3)(x +5) = 1/3 ⇒ 2x + 2/x² + 2x - 15 = 1/3 ⇒ x² + 2x - 15 = 6x + 6 ⇒ x² - 4x - 21 = 0 For the quadratic equatioRead more
Let the Rehman’s current age = x years
Therefore, 3 years ago, age = x -3 years
and, after 5 years, age = x + 5 years
according to questions,
1/x -3 + 1/x + 5 = 1/3
See less⇒ (x +5) + (x – 3)/(x – 3)(x +5) = 1/3
⇒ 2x + 2/x² + 2x – 15 = 1/3
⇒ x² + 2x – 15 = 6x + 6
⇒ x² – 4x – 21 = 0
For the quadratic equation x² – 4x – 21 = 0, we have a = 1, b = – 4 and c = – 21.
Therefore, b² – 4ac = (-4)² – 4 × 1 × (-21) = 16 + 84 = 100 > 0
Hence, x = 4±√100/2 = 4±10/2 [As x = (-b±√(b²-4ac))/2a]
Either x = 4+10/2 = 14/2 = 7 or x = 4 -10/2 = – 6/2 = – 3
Age of person can’t be negative, so Rehman’s age = 7 years.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square: 2x² + x + 4 = 0
2x² + x + 4 = 0 dividing both sides by 2 x² + 1/2 x = - 2 Adding [1/2(1/2)]² on both the sides, we get x² + 1/2 x + (1/4)² = - 2 + (1/4)² [ As x = (-b±√b²- 4ac)/2a] ⇒ (x+1/4)² = - 2 + 1/16 ⇒ (x + 1/4)² = (-32 + 1)/16 ⇒ (x + 1/4)² = - 31/16 < 0 we know that the square of any real number can not beRead more
2x² + x + 4 = 0
See lessdividing both sides by 2
x² + 1/2 x = – 2
Adding [1/2(1/2)]² on both the sides, we get
x² + 1/2 x + (1/4)² = – 2 + (1/4)² [ As x = (-b±√b²- 4ac)/2a]
⇒ (x+1/4)² = – 2 + 1/16 ⇒ (x + 1/4)² = (-32 + 1)/16
⇒ (x + 1/4)² = – 31/16 < 0
we know that the square of any real number can not be negative or (x + 1/4)² can not be negative. Hence, this quadratic equation has no real roots.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square: 4x² + 4√3x + 3 = 0
4x² + 4√3x + 3 = 0 For the quadratic equation 4x² + 4√3x + 3 = 0, we have a = 2, b = 1, and c = 4. Therefore, b² - 4ac = (4√3)² - 4 × 4 × 3 = 48 - 48 = 0 Hence, x = (-4√3±√0)/8 = -4√3/8 = - √3/2 [As, x = (-b±√b²-4ac)/2a] or x = -√3/2 Hence, the roots of quadratic equation are - √3/2 and - √3/2.
4x² + 4√3x + 3 = 0
See lessFor the quadratic equation 4x² + 4√3x + 3 = 0, we have a = 2, b = 1, and c = 4.
Therefore, b² – 4ac = (4√3)² – 4 × 4 × 3 = 48 – 48 = 0
Hence, x = (-4√3±√0)/8 = -4√3/8 = – √3/2 [As, x = (-b±√b²-4ac)/2a]
or x = -√3/2
Hence, the roots of quadratic equation are – √3/2 and – √3/2.
Find the roots of the following equations: 1/x + 4 – 1/x – 7 = 11/30, x ≠ – 4, 7
1/x + 4 - 1/x - 7 = 11/30, x ≠ – 4, 7 ⇒ (x -7) - (x +4)/(x + 4)(x - 7) = 11/30 ⇒ - 11/x² - 3x - 28 = 11/30 ⇒ x² - 3x - 28 = - 30 ⇒ x² - 3x + 2 = 0 For the quadratic equation x² - 3x + 2 = 0,we have a = 1, b = - 3, c = 1. Therefore, b² - 4ac = (-3)² - 4 × 1 × 2 = 9 - 8 = 1 > Hence, x = (3±√1)/2 =Read more
1/x + 4 – 1/x – 7 = 11/30, x ≠ – 4, 7
⇒ (x -7) – (x +4)/(x + 4)(x – 7) = 11/30
⇒ – 11/x² – 3x – 28 = 11/30
⇒ x² – 3x – 28 = – 30
⇒ x² – 3x + 2 = 0
For the quadratic equation x² – 3x + 2 = 0,we have a = 1, b = – 3, c = 1.
Therefore, b² – 4ac = (-3)² – 4 × 1 × 2 = 9 – 8 = 1 >
Hence, x = (3±√1)/2 = (3±1)/2 [As x = (-b±√b²-4ac)/2a
Either, x = 3+1/2 = 4/2 = 2 or x = 3-1/2 = 2/2 = 1
Hence, the roots of the quadratic equation are 2 and 1.
See less