2x² – 6x + 3 = 0 The given equations is of the form ax² + bx + c = 0 in which a = 2, b = - 6, c = 3. Therefore, D = b² - 4ac = (-6)² - 4 × 2 × 3 = 36 - 24 12 > 0 So, the roots of quadratic equation are real and unequal. Hence, x = 6±√12/4 = 6±3√3/4 = 3±√3/2 [As x = -b±√b² - 4ac/2a] Either x = 3±√Read more
2x² – 6x + 3 = 0
The given equations is of the form ax² + bx + c = 0 in which a = 2, b = – 6, c = 3.
Therefore, D = b² – 4ac = (-6)² – 4 × 2 × 3 = 36 – 24 12 > 0
So, the roots of quadratic equation are real and unequal.
Hence, x = 6±√12/4 = 6±3√3/4 = 3±√3/2 [As x = -b±√b² – 4ac/2a]
Either x = 3±√3/2 or x = 3-√3/2
Hence, the roots of the quadratic equation are 3+√3/2 and 3-√3/2.
Let the side of larger square = x m let the side of smaller square = y m According to question, x² + y² = 468 ...(i) Difference between parimeters, 4x - 4y = 24 ⇒ x - y = 6 ⇒ x = 6 + y ...(ii) Putting the value of x in equation (i), we get (y + 6)² + y² = 468 ⇒ y² + 12y + 36 + y² = 468 ⇒ 2y² + 12y -Read more
Let the side of larger square = x m
let the side of smaller square = y m
According to question, x² + y² = 468 …(i)
Difference between parimeters, 4x – 4y = 24
⇒ x – y = 6
⇒ x = 6 + y …(ii)
Putting the value of x in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468
⇒ 2y² + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18) (y – 12) = 0
⇒ (y + 18) = 0 or (y – 12) = 0
Either y = – 18 or y = 12
But, y ≠ – 18, as x is the side of square, which can’t be negative. So, y = 12
Hence, the side of smaller square = 12 m
Putting the value of y in equation (ii). we get
side of large square = x = y + 6 = 12 + 6 = 18 m.
Let, the speed of passenger train = x Km/h Therefore, the speed of express train = x + 11 km/h Distance travelled = 132 Km Time taken by passenger tarin t₁ = 133/x hours [As, time = distance/speed] Time taken by express train t₂ = 132/x+11 hours According to question, 132/x = 132/x + 11 = 1 ⇒ 132/xRead more
Let, the speed of passenger train = x Km/h
Therefore, the speed of express train = x + 11 km/h
Distance travelled = 132 Km
Time taken by passenger tarin t₁ = 133/x hours [As, time = distance/speed]
Time taken by express train t₂ = 132/x+11 hours
According to question,
132/x = 132/x + 11 = 1
⇒ 132/x = 132/x + 11 = 1
⇒ 132(x + 11) – 132x/x(x + 11) = 1
⇒ 132x + 1452 – 132x = x(x + 11)
⇒ 1452 = x² + 11x
⇒ x² + 11x – 1452 = 0
⇒ x² 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33(x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ (x + 44) = 0 or (x – 33) = 0
Either x = -44 or x = 33
But x ≠ -44, as x is the speed of train, which can’t be negative. So, x = 33
Hence, the speed of passenger train = 33Km/h
The speed of express train = 33 + 11 = 44Km/h
Let, the speed of train = x km/h Distance = 360 km Therefore, the time taken t₁ = 360/x hours [Because time = distance/speed] If the speed had been 5 km/h more, than the t₂ = 360/x+5 hours According to question, 360/x = 360/x+5 + 1 ⇒ 360/x - 360/x + 5 = 1 ⇒ 360(x + 5) - 360x/x(x+5) = 1 ⇒ 360x + 1800Read more
Let, the speed of train = x km/h
Distance = 360 km
Therefore, the time taken t₁ = 360/x hours [Because time = distance/speed]
If the speed had been 5 km/h more, than the t₂ = 360/x+5 hours
According to question,
360/x = 360/x+5 + 1
⇒ 360/x – 360/x + 5 = 1
⇒ 360(x + 5) – 360x/x(x+5) = 1
⇒ 360x + 1800 – 360x = x(x+5)
⇒ 1800 = x² + 5x
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ (x + 45) = 0 or (x – 40) = 0
Either x = – 45 or x = 40
But, x ≠ – 45, as x is the speed of train which can not be nagative.
So, x = 40
Hence, the speed of train is 40Km/h.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 2x² – 6x + 3 = 0
2x² – 6x + 3 = 0 The given equations is of the form ax² + bx + c = 0 in which a = 2, b = - 6, c = 3. Therefore, D = b² - 4ac = (-6)² - 4 × 2 × 3 = 36 - 24 12 > 0 So, the roots of quadratic equation are real and unequal. Hence, x = 6±√12/4 = 6±3√3/4 = 3±√3/2 [As x = -b±√b² - 4ac/2a] Either x = 3±√Read more
2x² – 6x + 3 = 0
See lessThe given equations is of the form ax² + bx + c = 0 in which a = 2, b = – 6, c = 3.
Therefore, D = b² – 4ac = (-6)² – 4 × 2 × 3 = 36 – 24 12 > 0
So, the roots of quadratic equation are real and unequal.
Hence, x = 6±√12/4 = 6±3√3/4 = 3±√3/2 [As x = -b±√b² – 4ac/2a]
Either x = 3±√3/2 or x = 3-√3/2
Hence, the roots of the quadratic equation are 3+√3/2 and 3-√3/2.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Let the side of larger square = x m let the side of smaller square = y m According to question, x² + y² = 468 ...(i) Difference between parimeters, 4x - 4y = 24 ⇒ x - y = 6 ⇒ x = 6 + y ...(ii) Putting the value of x in equation (i), we get (y + 6)² + y² = 468 ⇒ y² + 12y + 36 + y² = 468 ⇒ 2y² + 12y -Read more
Let the side of larger square = x m
See lesslet the side of smaller square = y m
According to question, x² + y² = 468 …(i)
Difference between parimeters, 4x – 4y = 24
⇒ x – y = 6
⇒ x = 6 + y …(ii)
Putting the value of x in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468
⇒ 2y² + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18) (y – 12) = 0
⇒ (y + 18) = 0 or (y – 12) = 0
Either y = – 18 or y = 12
But, y ≠ – 18, as x is the side of square, which can’t be negative. So, y = 12
Hence, the side of smaller square = 12 m
Putting the value of y in equation (ii). we get
side of large square = x = y + 6 = 12 + 6 = 18 m.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
Let, the speed of passenger train = x Km/h Therefore, the speed of express train = x + 11 km/h Distance travelled = 132 Km Time taken by passenger tarin t₁ = 133/x hours [As, time = distance/speed] Time taken by express train t₂ = 132/x+11 hours According to question, 132/x = 132/x + 11 = 1 ⇒ 132/xRead more
Let, the speed of passenger train = x Km/h
See lessTherefore, the speed of express train = x + 11 km/h
Distance travelled = 132 Km
Time taken by passenger tarin t₁ = 133/x hours [As, time = distance/speed]
Time taken by express train t₂ = 132/x+11 hours
According to question,
132/x = 132/x + 11 = 1
⇒ 132/x = 132/x + 11 = 1
⇒ 132(x + 11) – 132x/x(x + 11) = 1
⇒ 132x + 1452 – 132x = x(x + 11)
⇒ 1452 = x² + 11x
⇒ x² + 11x – 1452 = 0
⇒ x² 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33(x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ (x + 44) = 0 or (x – 33) = 0
Either x = -44 or x = 33
But x ≠ -44, as x is the speed of train, which can’t be negative. So, x = 33
Hence, the speed of passenger train = 33Km/h
The speed of express train = 33 + 11 = 44Km/h
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Let, the speed of train = x km/h Distance = 360 km Therefore, the time taken t₁ = 360/x hours [Because time = distance/speed] If the speed had been 5 km/h more, than the t₂ = 360/x+5 hours According to question, 360/x = 360/x+5 + 1 ⇒ 360/x - 360/x + 5 = 1 ⇒ 360(x + 5) - 360x/x(x+5) = 1 ⇒ 360x + 1800Read more
Let, the speed of train = x km/h
Distance = 360 km
Therefore, the time taken t₁ = 360/x hours [Because time = distance/speed]
If the speed had been 5 km/h more, than the t₂ = 360/x+5 hours
According to question,
360/x = 360/x+5 + 1
⇒ 360/x – 360/x + 5 = 1
⇒ 360(x + 5) – 360x/x(x+5) = 1
⇒ 360x + 1800 – 360x = x(x+5)
See less⇒ 1800 = x² + 5x
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ (x + 45) = 0 or (x – 40) = 0
Either x = – 45 or x = 40
But, x ≠ – 45, as x is the speed of train which can not be nagative.
So, x = 40
Hence, the speed of train is 40Km/h.