SEBA Class 10 Maths Chapter 4 Quadratic Equation

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SEBA Class 10 Maths Chapter 4 Quadratic Equation

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Quadratic Equation

Chapter: 4

Exercise 4.1

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2 (x – 3)

Ans: (x + 1)² = 2(x – 3) 

⇒ x²  + 2x + 1 = 2x – 6

⇒ x²  + 7 = 0 

⇒ x²  + 0.x + 7 = 0 

It is of the form ax² + bx + c = 0 

∴ The given equation is a quadratic equation. 

(ii)  x² – 2x = (-2) (3 – x)

Ans: x² – 2x = (-2)(3 – x)

⇒ x² – 2x = -6 + 2x

⇒ x² – 4x + 6 = 0

It is of the form ax² + bx + c = 0

∴ The given equation is quadratic.

(iii) (x – 2) (x + 1) =  (x – 1) (x + 3)

Ans: (x – 2) (x + 1) = (x – 1) (x + 3) 

⇒ x² + x – 2x – 2 = x² + 3x – x – 3

⇒ x² – x – 2 = x² + 2x – 2x – 3

⇒ -x – 2x – 2 + 3 = 0

⇒ -3x + 1 = 0 or 3x – 1 = 0

It is not of the form ax² + bx + c = 0

∴ The given equation is not quadratic equation.

(iv) (x – 3) (2x + 1) = x(x + 5)

Ans: (x – 3) (2x + 1) = x(x + 5)

⇒ 2 + x – 6x – 3 = x² + 5x

⇒ 2x² – 5x – 3 = x² + 5x

⇒ 2x² – x² – 5x – 5x – 3 = 0

⇒ x² – 10x – 3 = 0 It is a quadratic equation.

(v) (2x – 1) (x – 3) = (x + 5) (x – 1)

Ans: (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x² – 6x – x + 3 = x² + 5x – 5

⇒ 2x² – 7x + 3 = x² + 5x – 5

⇒ x²– 12x + 8 = 0

It is a quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

Ans: x² + 3x + 1 = (x – 2)²

⇒ x² + 3x + 1 = x² – 4x + 4

⇒ 3x + 4x + 1 – 4 = 0

⇒ 7x – 3 = 0 It is a linear equation and not a quadratic equation.

(vii) (x + 2)³ = 2x (x² – 1 )

Ans: (x + 2)³ = 2x(x² – 1)

⇒ x³ + 3x² × 2 + 3x + 2² + 2³ = 2x³ – 2x

⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x

⇒ -x³ + 6x² + 14x + 8 = 0

⇒ x³ – 6x² – 14x + 8 = 0

It is a cubic equation and not a quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

Ans: x³ – 4x² – x + 1 = (x – 2)³

⇒ x³ – 4x² – x + 1 = x³ + 3(x)(-2) (x – 2) – 2³

⇒ x³ – 4x² – x + 1 = x³ – 6x (x – 2) – 8

⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8

⇒ -4x² + 6x² – x – 12x + 1 + 8 = 0

⇒ 2x² – 13x + 9 = 0

It is a quadratic equation.

2. Represent the following situations in the form of quadratic equation:

(i) 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.

Ans: Let breadth be = x metres

Then length = (2x + 1) metres

∴ x × (2x + 1) = 528 (Area of the plot)

Or 2x² + x – 528 = 0

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Ans: Let the positive consecutive integers be x and x + 1

We are given that x × (x + 1) = 306 

⇒ x² + x – 306 = 0

(iii) Ram’s mother is 26 years older than him. The product of their ages (in years) 3 years from will be 360. We would like to find Ram’s present age.

Ans: Ram’s present age = x years

Then present age of Ram’s mother = (x + 26) years.

After three years, its is given that,

(x + 3) × {(x + 26 ) + 3} = 360

(x + 3) × (x + 29 ) = 360

X² + 32x + 87 = 360

x² + 32x + 87 – 360 = 0

x² + 32x – 273 = 0 

(iv) A train travels a distance of 480 km. at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Ans: Let the speed of the train be x km/hr

3. Choose the correct Option: For what value of p the equation (p – 2) x² + 3x + 5 = 0 can not be quadratic?

(a) 1

(b) 2

(c) -2

(d) 0

Ans. (b) 2

4. Which of the following are quadratic equations?

(i) (x + 1)² = 2(x – 4)

(ii) (x – 3)(x + 1) = (x + 2)(x – 3)

(iii) (x – 2)² + 1 = 2x – 4

(iv) x(x + 3) + 7 = (x + 2)(x – 2)

Choose the correct option.

(a) (i) and (iv)

(b) (i) and (ii)

(c) (i) and (iii)

(d) (ii) and (iv)

Ans. (c) (i) and (iii)

Exercise 4.2

1. Find the roots of the following quadratic equations Factorisation:

(i) x² – 3x – 10 = 0

Ans: x² – 3x – 10 = 0 

⇒ x² – 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5) (x + 2) = 0

⇒ (x – 5) = 0 or (x + 2) = 0

⇒ x = 5 or x = -2

Hence, the two roots are 5 and -2

(ii) 2x² + x – 6 = 0

Ans: 2x² + x – 6 = 0

⇒2x² + 4x – 3x – 6 = 0

⇒ 2x(x + 2) – 3(x + 2) = 0

 ⇒ (x + 2) (2x – 3) = 0

⇒ x + 2 = 0 or 2x – 3 = 0

⇒ x = -2 or x = 3/2

Hence the roots are -2 and 3/2

(iii) √2x² + 7x + 5√2 = 0

Ans: √2x² + 7x + 5√2 = 0

⇒ √2x² + 5x + 2x + 5√2 = 0

⇒ x(√2x + 5) + √2(√2x + 5) = 0

⇒ (x + √2) (√2x + 5) = 0

⇒ x = -√2 or x = -5/√2

(iv) 2x² – x + 1/8 = 0

Ans: 2x² – x + 1/8 = 0

⇒ 16x² – 8x + 1 = 0

⇒ (4x – 1)² = 0

⇒ Both roots are given by 4x – 1 = 0 i.e. x = 1/4

Hence, the roots are 1/4, 1/4

(v) 100x² – 20x + 1 = 0 

Ans: 100x² – 20x + 1 = 0

⇒ (10x – 1)² = 0

⇒ Both roots are given by 10x – 1 = 0 i.e. x = 1/10

Hence, the two roots are 1/10, 1/10

(vi) 2x² – 7x + 6 = 0

Ans: 2x² − 7x + 6 = 0

⇒ 2x² − 4x − 3x + 6 = 0

⇒ 2x(x − 2) − 3(x − 2) = 0

⇒ (x − 2)(2x − 3) = 0

Either, x − 2 = 0 or 2x − 3 = 0

⇒ x = 2  ⇒ x = 3/2

∴ The roots are 2, 3/2

(vii) x² – 10x – 96 = 0

Ans: x² − 10x − 96 = 0

⇒ x² − 16x + 6x − 96 = 0

⇒ x(x − 16) + 6(x − 16) = 0

⇒ (x − 16)(x + 6) = 0

Either x − 16 = 0 or x + 6 = 0

⇒ x = 16  ⇒ x = −6

∴ The roots are 16, −6.

(viii) √3x² + 10x + 7√3 = 0

Ans: √3x² + 10x + 7√3 = 0

⇒ √3x² + 7x + 3x + 7√3 = 0

⇒ x(√3x + 7) + √3(√3x + 7) = 0

⇒ (√3x + 7)(x + √3) = 0

Either √3x + 7 = 0 or x + √3 = 0

⇒ x = −7/√3  ⇒ x = −√3

∴ The roots are −7/√3, −√3

(ix) x² + 2√2x + 2 = 0

Ans: x² + 2√2x + 2 = 0

⇒ (x)² + 2 × √2 × 1x + (√2)² = 0

⇒ (x + √2)² = 0

⇒ (x + √2)(x + √2) = 0

Either x + √2 = 0 or x + √2 = 0

⇒ x = −√2 ⇒ x = −√2

∴ The roots are −√2 and −√2

(x) 14x + 5 – 3x² = 0

Ans: 14x + 5 − 3x² = 0

⇒ + 3x² − 14x − 5 = 0

⇒ 3x² − 15x + x − 5 = 0

⇒ 3x(x − 5) + 1(x − 5) = 0

⇒ (x − 5)(3x + 1) = 0

Either x − 5 = 0 or 3x + 1 = 0

⇒ x = 5 ⇒ x = −1/3

∴ The roots are 5, −1/3.

2. Represent the following situations mathematically:

(i) 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 324. We would like to find out how many marbles they had to start with.

Ans: The quadratic equation is x² – 45x + 324 = 0 

Where x is the number of marbles with John. 

The equation can be rewritten as

X² – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 9) (x – 36) = 0 

⇒ x = 9 or 36

Therefore, either john had 9 marbles and jivanti had 36 marbles or vice versa

(ii) 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 Rs. 750. We would like to find out the number of toys produced on that day.

3. Find the numbers whose sum is 27 and product is 182.

4. Find two consecutive positive integers, sum of whose squares is 365.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm., find the other two sides.

Ans: In ∆ΑΒC, base BC = x cm.

and altitude AC = (x – 7) cm. 

∠ACB = 90°, AB = 13cm.

By Pythagoras Theorem, we have, 

BC² + AC² = AB² 

⇒ x² + (x – 7)² = 13²

⇒ x² + x² – 14x + 49 = 169

⇒ 2x² – 14x – 120 = 0 

⇒ x² – 7x – 60 = 0

⇒ x² – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12) (x + 5) = 0

⇒ x – 5 or x = 12

Sides are not negative.

∴ x = 12  ∴ BC = 12 cm and AC = 5 cm. 

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day 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 Rs. 90, find the number of articles produced and the cost of each articles.

Ans: Let the number of articles produced in a day (on a particular day )= x 

Cost of production of one articles = Rs. (2x + 3)

We are given that, x × (2x + 3) = 90

⇒ 2x² + 3x – 90 = 0

⇒ 2x² + 15x – 12x – 90 = 0

⇒ x(2x + 15) – 6(2x + 15) = 0

⇒ (2x + 15) (x – 6) = 0

Now, 2x + 15 = 0

⇒ 2x = -15 ⇒ x = -15/2 (not exist)

Or, x – 6 = 0 ⇒ x = 6

Hence, the number of articles = 6

Cost of production of one article = Rs.(2 × 6 + 3) = Rs. 15

7. Choose the correct option: If x² – 2px + p² = 0 then the value of p/x is

(a) 0

(b) -1

(c) 1

(d) 2

Ans: (c) 1

8. A student has done through the following steps to find the roots of the equation. x² – 3x – 10 = 0 by factorization method

Step 1 : x² – 3x – 10 = 0

Step 2 : x² – 5x – 2x – 10 = 0

Step 3 : x(x – 5) – 2(x – 5) = 0

Step 4 : (x – 5) (x – 2) = 0

Step 5 : x = 5 and x = 2

In which step did the student make the first mistake?

(a) step: 2

(b) step: 3

(c) step: 4

(d) step: 5

Ans: (a) step: 2

9. Find the sum and product of the roots of the quadratic equation

2x² − 9x + 4 = 0

Ans: Standard form of quadratic equation is ax² + bx + c = 0

Here, a = 2, b = −9, c = 4

Sum of roots (α + β) = −b/a

⇒ α + β = −(−9)/2 = 9/2

Product of roots (αβ) = c/a

⇒ αβ = 4/2 = 2

∴ Sum of roots = 9/2 , Product of roots = 2

10. If one root of the quadratic equation 2x² + kx − 6 = 0 is 2, then find the value of k. Also find the other root.

Ans: Given, one root is x = 2

By substituting x = 2 in 2x² + kx − 6 = 0,

2(2²) + k(2) − 6 = 0

⇒ 2 × 4 + 2k − 6 = 0

⇒ 8 + 2k − 6 = 0

⇒ 2k + 2 = 0

⇒ 2k = −2

⇒ k = −1

now equation becomes 2x² − x − 6 = 0

Product of roots (αβ) = c/a

⇒ −6/2

= −3

Now, α = 2. Let the other root be β.

Then 2.β = −3

⇒ α = −3/2

∴ The required value of k = −1 and the value of other root (β) = −3/2.