Class |
10 |

Chapter |
Quadratic Equations |

Subject |
Math |

Category |
Important Question Answer |

**Class 10 Math Chapter 4 Important Question Answer**

**Q1. Find the root of the quadratic equation by factorization method.**

**Ans**.

=

=

So, the roots of the equation are the values of x for which

= 0

Now,

x =

and

x =

Therefore, the roots of are , .

**Q****2. For what value of k the roots of the quadratic equation 2x ^{2} + kx + 3 = 0 are equal?**

**Ans**. Quadratic equation has eqal roots when discriminant is equal to zero.

Here a = 2, b = k, c = 3.

Therefore, Discriminant b^{2} – 4ac = k^{2} – 4×2×3 = k^{2} – 24

Comparing it with 0,

k^{2} – 24 = 0

k =

Required values of k are -2√6 and 2√6.

**Q****3. Sum of the roots of the quadratic polynomnial 7x ^{2} – 3x + 1 is __________.**

**Ans**. a = 7, b = -3

Sum of roots =

**Q****4. Factorise the quadratic equation 2x ^{2} + x – 6 = 0 into linear factors.**

**Ans**. We have,

2x^{2} + x – 6 = 0

2x^{2} + 4x – 3x – 6 = 0

2x(x + 2) -3(x + 2) = 0

(2x – 3)(x + 2) = 0

Therefore, required linear factors of equation 2x^{2} + x – 6 are (2x – 3)(x + 2).

**Q****5. The roots of the quadratic equation 6x ^{2} – x – 2 = 0 are __________.**

**Ans**. We have

6x^{2} – x – 2 = 0

6x^{2} – 4x + 3x – 2** =** 0

2x(3x – 2) + 1(3x – 2) = 0

(2x + 1)(3x – 2) = 0

On comparing

2x + 1 = 0 and 3x – 2 = 0

x = and x =

Therefore, the roots of 6x^{2} – x – 2 = 0 are and .

**Q****6. For what values of K, quadratic equation x ^{2} + Kx + 4 = 0 has equal roots?**

**Ans**. Quadratic equation has eqal roots when discriminant is equal to zero.

Here a = 1, b = k, c = 4.

Therefore, Discriminant b^{2} – 4ac = k^{2} – 4×1×4 = k^{2} – 16

Comparing it with 0,

k^{2} – 16 = 0

k = ±4

Required values of k are -4 and 4.

**Q****7. Find discriminant of quadratic equation x ^{2} + 7x – 60 = 0.**

**Ans**. Here a = 1, b = 7, c = -60.

Therefore, discriminant b^{2} – 4ac = 7^{2 }– 4×1×(-60) = 49 + 240 = 289

**Q****8. Semi perimeter of a rectangular garden, whose length is 4m more: than its width is 36m. Find the dimensions of the garden.**

**Ans**. let breadth of rectangle = x m.

Then length will be (x + 4) m.

According to question,

Semi perimeter of rectangle = length + breadth = x + 4 + x = 36

2x + 4 = 36

2x = 32

x = 16 m.

length = x + 4 = 16 + 4 = 20 m.

Therefore, required dimensions of garden are 20 m and 16 m.

**Q****9. Find the nature of roots of the following quadratic equation. If real root exist, then solve it: 2x ^{2} – 6x + 3 = 0**

**Ans**. Nature of roots are determined by the discriminant of quadratic equation.

Here a = 2, b = -6, c = 3.

Therefore, discriminant b^{2} – 4ac = (-6)2 – 4×2×3 = 36 – 24 = 12.

Hence, b^{2} – 4ac > 0, two distinct real roots exist.

Finding roots of quadratic equation using quadratic formula,

=

Therefore required roots of given quadratic equation are .

**Q1****0. Find two consecutive positive integers the sum of whose squares is 365.**

**Ans**. Let the consecutive positive integers be x and x + 1.

According to question,

x^{2} + (x + 1)2 = 365

x^{2} + x^{2} + 1 + 2x = 365

2x^{2} + 2x – 364 = 0

Dividing by 2, we get

x^{2 }+ x – 182 = 0

x^{2 }+ 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x – 13)(x + 14) = 0

On comparing, we get

x – 13 = 0 i.e x = 13

x + 14 = 0 i.e x = -14

As we have to find positive integers. Therfore, x = -14 is not possible.

Therefore, required two consecutive positive numbers are 13 and 14.

**Q****11. Find two numbers whose sum is 37 and product is 300.**

Ans. Let 1^{st} number be x.

Then 2^{nd} number will be (37 – x)

According to question,

x(37 – x) = 300

37x – x^{2} = 300

37x – x^{2} – 300 = 0

x^{2} – 37x + 300 = 0

x^{2} – 12x – 25x + 300 = 0

x(x – 12) -25(x – 12) = 0

(x – 25)(x – 12) = 0

x – 25 = 0 x – 12 = 0

x = 25 x = 12

If 1^{st} number is 25, then 2^{nd} number will be 12.

Similarly, if 1^{st} number is 12, then 2^{nd} number will be 25.

Therefore, two pairs (12, 25) and (25, 12) are made such that two numbers whose sum is 37 and product is 300.

**Q****12. Find two numbers whose sum is 27 and product is 182.**

**Ans**. Let 1^{st} number be x.

Then 2^{nd} number will be (27 – x)

According to question,

x(27 – x) = 182

27x – x^{2} = 182

27x – x^{2} – 300 = 0

x^{2} – 27x + 182= 0

x^{2} – 13x –14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 14)(x – 13) = 0

x – 14 = 0 x – 13 = 0

x = 14 x = 13

If 1^{st} number is 14, then 2^{nd} number will be 13.

Similarly, if 1^{st} number is 13, then 2^{nd} number will be 14.

Therefore, two pairs (13, 14) and (14, 13) are made such that two numbers whose sum is 27 and product is 182.

**Q1****3. Is it possible to design a rectangular mango grove whose length is twice its breadth and area is 800 m ^{2}? If so, find its length and breadth.**

Ans. Let the breadth of the mango grove be x.

The length of the mango grove will be 2x.

Area of the mango grove = (2*x*) (x)= 2x^{2}

2x^{2 }= 80

x^{2 }= 800/2 = 400

x^{2} ^{ }– 400 =0

Comparing the given equation with *ax*^{2} + *bx *+ *c* = 0, we get

*a* = 1, *b* = 0, *c* = 400

As we know, discriminant = *b*^{2} – 4*ac*

=> (0)^{2} – 4 × (1) × ( – 400) = 1600

Here, *b*^{2} – 4*ac* > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

*x** *= ±20

As we know, the value of length cannot be negative.

Therefore, the breadth of the mango grove = 20 m.

Length of the mango grove = 2 × 20 = 40 m

**Q1****4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.**

**Ans**. Let’s the present age of one friend is x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of first friend = (*x* – 4) years

Age of second friend = (20 – *x* – 4) = (16 –* x*) years

As per the given question, we can write,

(*x* – 4) (16 – *x*) = 48

16*x – x*^{2} – 64 + 4*x* = 48

* – x*^{2} + 20*x – *112 = 0

*x*^{2} – 20*x + *112 = 0

Comparing the equation with *ax*^{2} + *bx *+ *c* = 0, we get

*a* = *1*, *b* = -2*0* and *c* = 112

Discriminant = *b*^{2} – 4*ac*

*= (-*20*)*^{2} – 4 × 112

= 400 – 448 = -48

*b*^{2} – 4*ac *< 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist .

**Q****15. One side of a rectangle exceeds its other side by 2cm. If its area is 195 cm ^{2} , then determine the sides of the rectangle.**

**Ans**. Let breadth of rectangle be x cm.

Then, length of triangle will be x + 2 cm.

According to question,

Area of rectangle = x(x + 2) = 195

x^{2} + 2x = 195

x^{2} + 2x – 195 = 0

x^{2} + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13)(x + 15) = 0

x – 13 = 0 or x + 15 = 0

x = 13 x = -15

As side of rectangle can’t be negative.

Therefore, x = 13 is true.

So, required sides of rectangle are 13 cm and 15 cm.

Also Read |
Class 10 Math NCERT Solution |

Also Read |
Class 10 Important Questions [Latest] |