Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.
JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.1
Question 1.
Check whether the following are quadratic equations:
1. (x + 1)2 = 2(x – 3)
2. x2 – 2x = (-2)(3 – x)
3. (x – 2)(x + 1) = (x – 1)(x + 3)
4. (x – 3)(2x + 1) = x(x + 5)
5. (2x – 1)(x – 3) = (x + 5)(x – 1)
6. x2 + 3x + 1 = (x – 2)2
7. (x + 2) = 2x(x2 – 1)
8. x3 – 4x2 – x + 1 = (x – 2)3
Solution:
1. Here, LHS = (x + 1)2 = x2 + 2x + 1 and
RHS = 2(x – 3) = 2x – 6.
Hence, (x + 1)2 = 2(x – 3) can be rewritten as x2 + 2x + 1 = 2x – 6
∴ x2 + 2x + 1 – 2x + 6 = 0
∴ x2 + 7 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = 0, c = 7)
Hence, the given equation is a quadratic equation.
2. Here, RHS = (-2)(3 – x) = -6 + 2x.
Hence, x2 – 2x = (-2) (3 – x) can be rewritten as
x2 – 2x = -6 + 2x
∴ x2 – 4x + 6 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = -4, c = 6)
Hence, the given equation is a quadratic equation.
3. Here, LHS = (x – 2) (x + 1) = x2 – x – 2 and
RHS = (x – 1)(x + 3) = x2 + 2x – 3.
Hence, (x – 2)(x + 1)(x – 1)(x + 3) be rewritten as
x2 – x – 2 = x2 + 2x – 3
∴ -3x + 1 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
4. Here, LHS = (x – 3) (2x + 1) = 2x2 – 5x – 3
and RHS = x(x + 5) = x2 + 5x.
Hence, (x – 3) (2x + 1) = x(x + 5) can be rewritten as
2x2 – 5x – 3 = x2 + 5x
∴ x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = -10, c = -3)
Hence, the given equation is a quadratic equation.
5. Here, LHS = (2x – 1)(x – 3) = 2x2 – 7x + 3 and
RHS = (x + 5) (x – 1) = x2 + 4x – 5.
Hence, the given equation can be rewritten as
2x2 – 7x + 3 = x2 + 4x – 5
∴ x2 – 11x + 8 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = -11, c = 8)
Hence, the given equation is a quadratic equation.
6. Here, RHS = (x – 2)2 = x2 – 4x + 4
Hence, the given equation can be rewritten as
x2 + 3x + 1 = x2 – 4x + 4
∴ 7x – 3 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
7. Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8 and
RHS = 2x (x2 – 1) = 2x3 – 2x.
Hence, the given equation can be rewritten as
x3 + 6x2 + 12x + 8 = 2x3 – 2x
∴ -x3 + 6x2 + 14x + 8 = 0
It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
8. Here, RHS = (x – 2)3 = x3 – 6x2 + 12x – 8.
Hence, the given equation can be rewritten as
x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
2x2 – 13x + 9 = 0
It is of the form ax2 + bx + c = 0.
(a = 2, b = -13, c = 9)
Hence, the given equation is a quadratic equation.
Question 2.
Represent the following situations in the form of quadratic equations:
1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth (in metres) of the rectangular plot be x.
Then, the length (in metres) of the rectangular plot is 2x + 1.
Area of the rectangular plot = Length × Breadth
∴ 528 = (2x + 1) × x
(∵ Area is given to be 528 m2)
∴ 528 = 2x2 + x
∴ 2x2 + x – 528 = 0 is the required quadratic equation to find the length (2x + 1 m) and breadth (x m) of the rectangular plot.
2. The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let the two consecutive positive integer be x and x + 1.
Then, their product = x(x + 1) = x2 + x.
This product is given to be 306.
∴ x2 + x = 306
∴ x2 + x – 306 = 0 is the required quadratic equation to find the consecutive positive integers x and x + 1.
3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:
Let Rohan’s present age (in years) be x.
Then, his mother’s present age (in years) = x + 26.
3 years from now, Rohan’s age (in years) will be x + 3 and his mother’s age (in years) will be x + 29.
The product of their ages (in years) 3 years from now is given to be 360.
Hence, (x + 3)(x + 29) = 360
∴ x2 + 32x + 87 – 360 = 0
∴ x2 + 32x – 273 = 0 is the required quadratic equation to find the present ages (in years) of Rohan (x) and his mother (x + 26).
4. 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.
Solution:
Let the usual uniform speed of the train be x km/h.
Now, Time = \(\frac{\text { distance }}{\text { speed }}\)
∴ Time required to cover 480 km distance at usual speed = t1 = \(\frac{480}{x}\) hours
If the speed is 8 km/hour less, the new speed would be (x – 8) km/hour.
∴ Time required to cover 480 km distance at new speed = t2 = \(\frac{480}{x-8}\) hours
Now, the time required at new speed is 3 hours more than the usual time.
∴ t2 = t1 + 3
∴ \(\frac{480}{x-8}=\frac{480}{x}+3\)
∴ 480x = 480(x – 8) + 3x(x – 8)
(Multiplying by x(x – 8))
∴ 480x = 480x – 3840 + 3x2 – 24x
∴ 0 = 3x2 – 24x – 3840
∴ x2 – 8x – 1280 = 0 is the required quadratic equation to find the usual speed (x km/h) of the train.