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Maths Quadratic Equations Test.19

**Quiz Instructions:**

- There will be 20 multiple choice question in this online test.
- Answer of the questions will change randomly each time you start this test.
- Practice this test at least 3 times if you want to secure High Marks.
- At the End of the Test you can see your Test score and Rating.

1 / 20

**Find the roots of quadratic equation: 3x ^{2} – 7x – 6 = 0?**

3×2 – 9x + 2x – 6 = 0

3x(x – 3) + 2(x – 3) = 0

(x – 3)(3x + 2) = 0 => x = 3, -2/3

3×2 – 9x + 2x – 6 = 0

3x(x – 3) + 2(x – 3) = 0

(x – 3)(3x + 2) = 0 => x = 3, -2/3

2 / 20

**Find the roots of quadratic equation: x ^{2} + x – 42 = 0?**

Explanation:

x2 + 7x – 6x + 42 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7)(x – 6) = 0 => x = -7, 6

Explanation:

x2 + 7x – 6x + 42 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7)(x – 6) = 0 => x = -7, 6

3 / 20

**Find the roots of quadratic equation: 2x ^{2} + 5x + 2 = 0?**

Explanation:

2×2 + 4x + x + 2 = 0

2x(x + 2) + 1(x + 2) = 0

(x + 2)(2x + 1) = 0 => x = -2, -1/2

Explanation:

2×2 + 4x + x + 2 = 0

2x(x + 2) + 1(x + 2) = 0

(x + 2)(2x + 1) = 0 => x = -2, -1/2

4 / 20

**I. a ^{2} – 13a + 42 = 0,II. b^{2} – 15b + 56 = 0 to solve both the equations to find the values of a and b?**

Explanation:

I. a2 – 13a + 42 = 0

=>(a – 6)(a – 7) = 0 => a = 6, 7

II. b2 – 15b + 56 = 0

=>(b – 7)(b – 8) = 0 => b = 7, 8

=>a ≤ b

Explanation:

I. a2 – 13a + 42 = 0

=>(a – 6)(a – 7) = 0 => a = 6, 7

II. b2 – 15b + 56 = 0

=>(b – 7)(b – 8) = 0 => b = 7, 8

=>a ≤ b

5 / 20

**I. a ^{3} – 988 = 343,II. b^{2} – 72 = 49 to solve both the equations to find the values of a and b?**

Explanation:

a3 = 1331 => a = 11

b2 = 121 => b = ± 11

a ≥ b

Explanation:

a3 = 1331 => a = 11

b2 = 121 => b = ± 11

a ≥ b

6 / 20

**I. 9a ^{2} + 18a + 5 = 0,II. 2b^{2} + 13b + 20 = 0 to solve both the equations to find the values of a and b?**

Explanation:

I. 9a2 + 3a + 15a + 5 = 0

=>(3a + 5)(3a + 1) = 0 => a = -5/3, -1/3

II. 2b2 + 8b + 5b + 20 = 0

=>(2b + 5)(b + 4) = 0 => b = -5/2, -4

a is always more than b.

a > b.

Explanation:

I. 9a2 + 3a + 15a + 5 = 0

=>(3a + 5)(3a + 1) = 0 => a = -5/3, -1/3

II. 2b2 + 8b + 5b + 20 = 0

=>(2b + 5)(b + 4) = 0 => b = -5/2, -4

a is always more than b.

a > b.

7 / 20

**I. x ^{2} + 11x + 30 = 0,II. y^{2} + 15y + 56 = 0 to solve both the equations to find the values of x and y?**

Explanation:

I. x2 + 6x + 5x + 30 = 0

=>(x + 6)(x + 5) = 0 => x = -6, -5

II. y2 + 8y + 7y + 56 = 0

=>(y + 8)(y + 7) = 0 => y = -8, -7

=> x > y

Explanation:

I. x2 + 6x + 5x + 30 = 0

=>(x + 6)(x + 5) = 0 => x = -6, -5

II. y2 + 8y + 7y + 56 = 0

=>(y + 8)(y + 7) = 0 => y = -8, -7

=> x > y

8 / 20

**I. x ^{2} + 3x – 18 = 0,II. y^{2} + y – 30 = 0 to solve both the equations to find the values of x and y?**

Explanation:

I. x2 + 6x – 3x – 18 = 0

=>(x + 6)(x – 3) = 0 => x = -6, 3

II. y2 + 6y – 5y – 30 = 0

=>(y + 6)(y – 5) = 0 => y = -6, 5

No relationship can be established between x and y.

Explanation:

I. x2 + 6x – 3x – 18 = 0

=>(x + 6)(x – 3) = 0 => x = -6, 3

II. y2 + 6y – 5y – 30 = 0

=>(y + 6)(y – 5) = 0 => y = -6, 5

No relationship can be established between x and y.

9 / 20

**I. x ^{2} + 9x + 20 = 0,II. y^{2} + 5y + 6 = 0 to solve both the equations to find the values of x and y?**

I. x2 + 4x + 5x + 20 = 0

=>(x + 4)(x + 5) = 0 => x = -4, -5

II. y2 + 3y + 2y + 6 = 0

=>(y + 3)(y + 2) = 0 => y = -3, -2

= x < y.

I. x2 + 4x + 5x + 20 = 0

=>(x + 4)(x + 5) = 0 => x = -4, -5

II. y2 + 3y + 2y + 6 = 0

=>(y + 3)(y + 2) = 0 => y = -3, -2

= x < y.

10 / 20

**I. x ^{2} – x – 42 = 0,II. y^{2} – 17y + 72 = 0 to solve both the equations to find the values of x and y?**

Explanation:

I. x^{2} – 7x + 6x – 42 = 0

=> (x – 7)(x + 6) = 0 => x = 7, -6

II. y^{2} – 8y – 9y + 72 = 0

=> (y – 8)(y – 9) = 0 => y = 8, 9

=> x < y

Explanation:

I. x^{2} – 7x + 6x – 42 = 0

=> (x – 7)(x + 6) = 0 => x = 7, -6

II. y^{2} – 8y – 9y + 72 = 0

=> (y – 8)(y – 9) = 0 => y = 8, 9

=> x < y

11 / 20

**I. x ^{2} + 5x + 6 = 0,II. y^{2} + 9y +14 = 0 to solve both the equations to find the values of x and y?**

Explanation:

I. x^{2} + 3x + 2x + 6 = 0

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

II. y^{2} + 7y + 2y + 14 = 0

=> (y + 7)(y + 2) = 0 => y = -7 or -2

No relationship can be established between x and y.

Explanation:

I. x^{2} + 3x + 2x + 6 = 0

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

II. y^{2} + 7y + 2y + 14 = 0

=> (y + 7)(y + 2) = 0 => y = -7 or -2

No relationship can be established between x and y.

12 / 20

**I. a ^{2} + 8a + 16 = 0,II. b^{2} – 4b + 3 = 0 to solve both the equations to find the values of a and b?**

I. (a + 4)2 = 0 => a = -4

II.(b – 3)(b – 1) = 0

=> b = 1, 3 => a < b

I. (a + 4)2 = 0 => a = -4

II.(b – 3)(b – 1) = 0

=> b = 1, 3 => a < b

13 / 20

**(i). a ^{2} + 11a + 30 = 0,(ii). b^{2} + 6b + 5 = 0 to solve both the equations to find the values of a and b?**

Explanation:

I. (a + 6)(a + 5) = 0

=> a = -6, -5

II. (b + 5)(b + 1) = 0

=> b = -5, -1 => a ≤ b

Explanation:

I. (a + 6)(a + 5) = 0

=> a = -6, -5

II. (b + 5)(b + 1) = 0

=> b = -5, -1 => a ≤ b

14 / 20

**(i). a ^{2} – 9a + 20 = 0,(ii). 2b^{2} – 5b – 12 = 0 to solve both the equations to find the values of a and b?**

Explanation:

I. (a – 5)(a – 4) = 0

=> a = 5, 4

II. (2b + 3)(b – 4) = 0

=> b = 4, -3/2 => **a ≥ b**

Explanation:

I. (a – 5)(a – 4) = 0

=> a = 5, 4

II. (2b + 3)(b – 4) = 0

=> b = 4, -3/2 => **a ≥ b**

15 / 20

**(i). a ^{2} – 7a + 12 = 0,(ii). b^{2} – 3b + 2 = 0 to solve both the equations to find the values of a and b?**

Explanation:

I.(a – 3)(a – 4) = 0

=> a = 3, 4

II. (b – 2)(b – 1) = 0

=> b = 1, 2

=> **a > b**

Explanation:

I.(a – 3)(a – 4) = 0

=> a = 3, 4

II. (b – 2)(b – 1) = 0

=> b = 1, 2

=> **a > b**

16 / 20

**A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?**

Explanation:

Let the price of each note book be Rs.x.

Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.

Hence xy = 300

=> y = 300/x

(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy

=>5(300/x) – 10x – 50 = 0 => -150 + x2 + 5x = 0

multiplying both sides by -1/10x

=> x2 + 15x – 10x – 150 = 0

=> x(x + 15) – 10(x + 15) = 0

=> x = 10 or -15

As x>0, x = **10**

Explanation:

Let the price of each note book be Rs.x.

Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.

Hence xy = 300

=> y = 300/x

(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy

=>5(300/x) – 10x – 50 = 0 => -150 + x2 + 5x = 0

multiplying both sides by -1/10x

=> x2 + 15x – 10x – 150 = 0

=> x(x + 15) – 10(x + 15) = 0

=> x = 10 or -15

As x>0, x = **10**

17 / 20

**Find the quadratic equations whose roots are the reciprocals of the roots of 2x ^{2} + 5x + 3 = 0?**

The quadratic equation whose roots are reciprocal of 2x^{2} + 5x + 3 = 0 can be obtained by replacing x by 1/x.

Hence, 2(1/x)^{2} + 5(1/x) + 3 = 0

=> **3x ^{2} + 5x + 2 = 0**

The quadratic equation whose roots are reciprocal of 2x^{2} + 5x + 3 = 0 can be obtained by replacing x by 1/x.

Hence, 2(1/x)^{2} + 5(1/x) + 3 = 0

=> **3x ^{2} + 5x + 2 = 0**

18 / 20

**Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x ^{2} + 8x + 4 = 0?**

Explanation:

a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab

= [(a + b)2 – 2ab]/ab

a + b = -8/1 = -8

ab = 4/1 = 4

Hence a/b + b/a = [(-8)2 – 2(4)]/4 = 56/4 = **14.**

Explanation:

a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab

= [(a + b)2 – 2ab]/ab

a + b = -8/1 = -8

ab = 4/1 = 4

Hence a/b + b/a = [(-8)2 – 2(4)]/4 = 56/4 = **14.**

19 / 20

**The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?**

Explanation:

Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.

(2x – 2)2 + (2x)2 + (2x + 2)2 = 1460

4×2 – 8x + 4 + 4×2 + 8x + 4 = 1460

12×2 = 1452 => x2 = 121 => x = ± 11

As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.

Required numbers are **20, 22, 24.**

Explanation:

Three consecutive even natural numbers be 2x – 2, 2x and 2x + 2.

(2x – 2)2 + (2x)2 + (2x + 2)2 = 1460

4×2 – 8x + 4 + 4×2 + 8x + 4 = 1460

12×2 = 1452 => x2 = 121 => x = ± 11

As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.

Required numbers are **20, 22, 24.**

20 / 20

**If a and b are the roots of the equation x ^{2} – 9x + 20 = 0, find the value of a^{2} + b^{2} + ab?**

Explanation:

a^{2} + b^{2} + ab = a^{2} + b^{2} + 2ab – ab

i.e., (a + b)^{2} – ab

from x^{2} – 9x + 20 = 0, we have

a + b = 9 and ab = 20. Hence the value of required expression (9)^{2} – 20 = **61.**

Explanation:

a^{2} + b^{2} + ab = a^{2} + b^{2} + 2ab – ab

i.e., (a + b)^{2} – ab

from x^{2} – 9x + 20 = 0, we have

a + b = 9 and ab = 20. Hence the value of required expression (9)^{2} – 20 = **61.**