Let the speed of train = x Km/h Total distance = 480 Km Therefore, time taken = 480/x hours If the speed had been 8 Km/h less, than time taken = 480/x-8 hours According to the questions, 480/x-8 - 480/x = 3 ⇒ 480x - 480(x - 8)/(x - 8)x = 3 ⇒ 480x - 480x + 3640 = 3(x - 8)x ⇒ 3640 = 3x² - 24x ⇒ 3x² -Read more
Let the speed of train = x Km/h
Total distance = 480 Km
Therefore, time taken = 480/x hours
If the speed had been 8 Km/h less, than time taken = 480/x-8 hours
According to the questions,
480/x-8 – 480/x = 3
⇒ 480x – 480(x – 8)/(x – 8)x = 3
⇒ 480x – 480x + 3640 = 3(x – 8)x
⇒ 3640 = 3x² – 24x
⇒ 3x² – 24x – 3640 = 0
Hence, the speed of train satisfies the quadratic equation 3x² – 24x – 3640 = 0.
Lets, the number of article = x Therefore, the cost of one article = 2x + 3 According to the 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 = - 15/2Read more
Lets, the number of article = x
Therefore, the cost of one article = 2x + 3
According to the 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, x as is number of articles.
Therefore, x = 6 and the cost of each article 2x + 3 = 2 × 6 + 3 = 15
Hence, the number of articles = 6 and the cost of each article is Rs 15.
Let the number of marbles John had be x. Then the number of marbles Jivanti had = 45 – x (Why?). The number of marbles left with John, when he lost 5 marbles = x – 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x Therefore, their product = (x – 5) (40 – x) = 4Read more
Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= – x2 + 45x – 200
So, – x2 + 45x – 200 = 124 (Given that product = 124)
i.e., – x2 + 45x – 324 = 0
i.e., x2 – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x2 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(x+1)² = 2(x - 3) simplifying the given equation, we got (x + 1)² = 2(x-3) ⇒ x² + 2x + 1 = 2x - 6 ⇒ x ² + 7 = 0 or x² +0x + 7 = 0 this is an equation of type ax² + bx + c = 0 Hence, the given equation is a quadratic equation. video explanation
(x+1)² = 2(x – 3)
simplifying the given equation, we got
(x + 1)² = 2(x-3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x ² + 7 = 0
or x² +0x + 7 = 0
this is an equation of type ax² + bx + c = 0
Hence, the given equation is a quadratic equation.
Represent the situations in the form of quadratic equations: A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Let the speed of train = x Km/h Total distance = 480 Km Therefore, time taken = 480/x hours If the speed had been 8 Km/h less, than time taken = 480/x-8 hours According to the questions, 480/x-8 - 480/x = 3 ⇒ 480x - 480(x - 8)/(x - 8)x = 3 ⇒ 480x - 480x + 3640 = 3(x - 8)x ⇒ 3640 = 3x² - 24x ⇒ 3x² -Read more
Let the speed of train = x Km/h
See lessTotal distance = 480 Km
Therefore, time taken = 480/x hours
If the speed had been 8 Km/h less, than time taken = 480/x-8 hours
According to the questions,
480/x-8 – 480/x = 3
⇒ 480x – 480(x – 8)/(x – 8)x = 3
⇒ 480x – 480x + 3640 = 3(x – 8)x
⇒ 3640 = 3x² – 24x
⇒ 3x² – 24x – 3640 = 0
Hence, the speed of train satisfies the quadratic equation 3x² – 24x – 3640 = 0.
Find the roots of the following quadratic equations by factorisation: √2x² + 7x + 5√2 = 0
√2x² + 7x + 5√2 = 0 solving the quadratic equation, we get √2x² + 7x + 5√2 = 0 ⇒ √2x² + 5x + 2x 5√2 = 0 ⇒ x(√2x + 5) +√2(√2x + 5) = 0 ⇒ (√2x + 5) (x + √2) = 0 ⇒ (√2x + 5) = 0 or (x + √2) = 0 Either x = - 5/√2 or = - √2 Hence, the roots of the given quadratic equation are - 5/√2 and - √2.
√2x² + 7x + 5√2 = 0
See lesssolving the quadratic equation, we get
√2x² + 7x + 5√2 = 0
⇒ √2x² + 5x + 2x 5√2 = 0
⇒ x(√2x + 5) +√2(√2x + 5) = 0
⇒ (√2x + 5) (x + √2) = 0
⇒ (√2x + 5) = 0 or (x + √2) = 0
Either x = – 5/√2 or = – √2
Hence, the roots of the given quadratic equation are – 5/√2 and – √2.
Represent the following situations mathematically: A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.
Lets, the number of article = x Therefore, the cost of one article = 2x + 3 According to the 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 = - 15/2Read more
Lets, the number of article = x
See lessTherefore, the cost of one article = 2x + 3
According to the 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, x as is number of articles.
Therefore, x = 6 and the cost of each article 2x + 3 = 2 × 6 + 3 = 15
Hence, the number of articles = 6 and the cost of each article is Rs 15.
Represent the following situations mathematically: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Let the number of marbles John had be x. Then the number of marbles Jivanti had = 45 – x (Why?). The number of marbles left with John, when he lost 5 marbles = x – 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x Therefore, their product = (x – 5) (40 – x) = 4Read more
Let the number of marbles John had be x.
See lessThen the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= – x2 + 45x – 200
So, – x2 + 45x – 200 = 124 (Given that product = 124)
i.e., – x2 + 45x – 324 = 0
i.e., x2 – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x2 – 45x + 324 = 0
which is the required representation of the problem mathematically.
Find the roots of the following quadratic equations by factorisation: 100 x² – 20x + 1 = 0
100 x² – 20x + 1 = 0 Solving the quadratic equation, we get, 100 x² – 20x + 1 = 0 ⇒ 100x² - 10x - 10x + 1 = 0 ⇒ 10x(10x - 1) - 10(10x - 1) = 0 ⇒ (10x -1) (10x - 1) = 0 ⇒ (10x - 1) = 0 or (10x - 1) = 0 Either x = 1/10 or x = 1/10 Hence, the roots of given quadratic equation are 1/10 and 1/10.
100 x² – 20x + 1 = 0
See lessSolving the quadratic equation, we get,
100 x² – 20x + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – 10(10x – 1) = 0
⇒ (10x -1) (10x – 1) = 0
⇒ (10x – 1) = 0 or (10x – 1) = 0
Either x = 1/10 or x = 1/10
Hence, the roots of given quadratic equation are 1/10 and 1/10.
Find the roots of the following quadratic equations by factorisation: 2x² – x +1/8 = 0
2x² – x + 1/8 = 0 Solving the quadratic equation, we got 16x² - 8x + 1 = 0 ⇒ 16x² - 4x - 4x + 1 = 0 ⇒ 4x(4x -1) - 1(4x -1) = 0 ⇒ (4x - 1) (4x -1) = 0 ⇒ (4x - 1) = 0 or (4x - 1) = 0 Either x = 1/4 or x = 1/4 Hence, the roots of given quadratic equation are 1/4 and 1/4.
2x² – x + 1/8 = 0
See lessSolving the quadratic equation, we got
16x² – 8x + 1 = 0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x -1) – 1(4x -1) = 0
⇒ (4x – 1) (4x -1) = 0
⇒ (4x – 1) = 0 or (4x – 1) = 0
Either x = 1/4 or x = 1/4
Hence, the roots of given quadratic equation are 1/4 and 1/4.
Check whether the following are quadratic equations : (x + 1)² = 2(x – 3)
(x+1)² = 2(x - 3) simplifying the given equation, we got (x + 1)² = 2(x-3) ⇒ x² + 2x + 1 = 2x - 6 ⇒ x ² + 7 = 0 or x² +0x + 7 = 0 this is an equation of type ax² + bx + c = 0 Hence, the given equation is a quadratic equation. video explanation
(x+1)² = 2(x – 3)
simplifying the given equation, we got
(x + 1)² = 2(x-3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x ² + 7 = 0
or x² +0x + 7 = 0
this is an equation of type ax² + bx + c = 0
Hence, the given equation is a quadratic equation.
video explanation
See less