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.
Let the smaller side = x m Therefore, hypotenuse = x + 60 m So, the longer side = x + 30 m According to the question, (x + 60)² = x² + (x + 30)² ⇒ x² + 120x +3600 = x² + x² + 60x + 900 ⇒ - x² + 60x + 2700 = 0 ⇒ x² - 60 - 2700 = 0 ⇒ x² - 90 + 30x - 2700 = 0 ⇒ x(x - 90) + 30 (x - 90) = 0 ⇒ (x - 90) (xRead more
Let the smaller side = x m
Therefore, hypotenuse = x + 60 m
So, the longer side = x + 30 m
According to the question, (x + 60)² = x² + (x + 30)²
⇒ x² + 120x +3600 = x² + x² + 60x + 900
⇒ – x² + 60x + 2700 = 0
⇒ x² – 60 – 2700 = 0
⇒ x² – 90 + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ (x – 90) = 0 or ( x + 30 ) = 0
Either x = 90 or x = – 30
But, x ≠ – 30 ,as x is side of field which can’t be negative.
Therefore, x = 90 and hence the smaller side = 90 m
So, the longer side = 90 + 30 = 120 m
Let the larger number = x Let the smaller number = y Therefore, y² = 8x According to the question, x² - y² = 180 ⇒ x² - 8x = 180 [ As, y² = 8x ] ⇒ x² - 8x - 180 = 0 ⇒ x² - 18x + 10x - 180 = 0 ⇒ x(x - 18) +10 (x -18) = 0 ⇒ (x - 18) (x + 10) = 0 ⇒ (x - 18) = 0 or, (x + 10) = 0 Either x = 18 or x = - 1Read more
Let the larger number = x
Let the smaller number = y
Therefore, y² = 8x
According to the question, x² – y² = 180
⇒ x² – 8x = 180 [ As, y² = 8x ]
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x(x – 18) +10 (x -18) = 0
⇒ (x – 18) (x + 10) = 0
⇒ (x – 18) = 0 or, (x + 10) = 0
Either x = 18 or x = – 10
But, x ≠ – 10, as x larger of the two numbers So, x = 18
Therefore, the larger number = 18
Hence the smaller number = y = √144 = 12
Let, the number of article = x Therefore, the cost of one article = 2x + 3 According to question, the total cost = x(2x + 3) = 90 ⇒ 2x² +3x = 90 ⇒ 2x² + 3x - 90 = 0 ⇒ 2x² + 15x - 12x - 90 = 0 ⇒ x(2x + 15) - 6 (2x + 15) = 0 ⇒ (2x + 15) (x - 6) = 0 ⇒ (2x + 15) = 0 or (x - 6) = 0 Either x = - 15/2 or xRead more
Let, the number of article = x
Therefore, the cost of one article = 2x + 3
According to question, the total cost = x(2x + 3) = 90
⇒ 2x² +3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6 (2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ (2x + 15) = 0 or (x – 6) = 0
Either x = – 15/2 or x = 6
But, x ≠ – 15/2, as x is a number of articles.
Therefore, x = 6 and the cost of each articles = 2x + 3 = 2 × 6 + 3 = 15
Hence, the number of articles = 6 and the cost of each article is Rs 15.
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 lessFind the roots of the following equations: x – 1/x = 3, x ≠ 0
x - 1/x = 3, x ≠ 0 ⇒ x² - 1 = 3x ⇒ x² - 3x - 1 = 0 For the quadratic equation x² - 3x - 1 = 0, we have a = 1, b = - 3 and c = - 1. Therefore, b² - 4ac = (-3)² - 4 × 1 × (-1) = 9 + 4 = 13 > 0 Hence, x = (3 ± √13)/2 [As x = (-b ±√(b²- 4ac))/2a] Either x = (3 + √13)/2 or x = (3 - √13)/2 Hence, the rRead more
x – 1/x = 3, x ≠ 0
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
For the quadratic equation x² – 3x – 1 = 0, we have a = 1, b = – 3 and c = – 1.
Therefore, b² – 4ac = (-3)² – 4 × 1 × (-1) = 9 + 4 = 13 > 0
Hence, x = (3 ± √13)/2 [As x = (-b ±√(b²- 4ac))/2a]
Either x = (3 + √13)/2 or x = (3 – √13)/2
See lessHence, the roots of the quadratic equation are (3 + √13)/2 and (3 – √13)/2.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Let the smaller side = x m Therefore, hypotenuse = x + 60 m So, the longer side = x + 30 m According to the question, (x + 60)² = x² + (x + 30)² ⇒ x² + 120x +3600 = x² + x² + 60x + 900 ⇒ - x² + 60x + 2700 = 0 ⇒ x² - 60 - 2700 = 0 ⇒ x² - 90 + 30x - 2700 = 0 ⇒ x(x - 90) + 30 (x - 90) = 0 ⇒ (x - 90) (xRead more
Let the smaller side = x m
See lessTherefore, hypotenuse = x + 60 m
So, the longer side = x + 30 m
According to the question, (x + 60)² = x² + (x + 30)²
⇒ x² + 120x +3600 = x² + x² + 60x + 900
⇒ – x² + 60x + 2700 = 0
⇒ x² – 60 – 2700 = 0
⇒ x² – 90 + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ (x – 90) = 0 or ( x + 30 ) = 0
Either x = 90 or x = – 30
But, x ≠ – 30 ,as x is side of field which can’t be negative.
Therefore, x = 90 and hence the smaller side = 90 m
So, the longer side = 90 + 30 = 120 m
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Let the larger number = x Let the smaller number = y Therefore, y² = 8x According to the question, x² - y² = 180 ⇒ x² - 8x = 180 [ As, y² = 8x ] ⇒ x² - 8x - 180 = 0 ⇒ x² - 18x + 10x - 180 = 0 ⇒ x(x - 18) +10 (x -18) = 0 ⇒ (x - 18) (x + 10) = 0 ⇒ (x - 18) = 0 or, (x + 10) = 0 Either x = 18 or x = - 1Read more
Let the larger number = x
See lessLet the smaller number = y
Therefore, y² = 8x
According to the question, x² – y² = 180
⇒ x² – 8x = 180 [ As, y² = 8x ]
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x(x – 18) +10 (x -18) = 0
⇒ (x – 18) (x + 10) = 0
⇒ (x – 18) = 0 or, (x + 10) = 0
Either x = 18 or x = – 10
But, x ≠ – 10, as x larger of the two numbers So, x = 18
Therefore, the larger number = 18
Hence the smaller number = y = √144 = 12
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was rupees 90, find the number of articles produced and the cost of each article.
Let, the number of article = x Therefore, the cost of one article = 2x + 3 According to question, the total cost = x(2x + 3) = 90 ⇒ 2x² +3x = 90 ⇒ 2x² + 3x - 90 = 0 ⇒ 2x² + 15x - 12x - 90 = 0 ⇒ x(2x + 15) - 6 (2x + 15) = 0 ⇒ (2x + 15) (x - 6) = 0 ⇒ (2x + 15) = 0 or (x - 6) = 0 Either x = - 15/2 or xRead more
Let, the number of article = x
See lessTherefore, the cost of one article = 2x + 3
According to question, the total cost = x(2x + 3) = 90
⇒ 2x² +3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6 (2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ (2x + 15) = 0 or (x – 6) = 0
Either x = – 15/2 or x = 6
But, x ≠ – 15/2, as x is a number of articles.
Therefore, x = 6 and the cost of each articles = 2x + 3 = 2 × 6 + 3 = 15
Hence, the number of articles = 6 and the cost of each article is Rs 15.