ML Aggarwal Quadratic Equations Exe-5.1 Class 10 ICSE Maths Solutions . We Provide Step by Step Answer of Exe-5.1 Questions for Quadratic Equations in One Variable as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Ch-5 Quadratic Equations in one Variable Exercise- 5.1 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-5 | Quadratic Equations |

Writer / Book | Understanding |

Topics | Solutions of Exe-5 |

Academic Session | 2024-2025 |

** Quadratic Equations in one Variable Exe-5.1**

(Class 10 ICSE ML Aggarwal Maths Solutions)** **

**Question -1. ****In each of the following, determine whether the given numbers are roots of the given equations or not.**

(i) x² – 5x + 6 = 0; 2, – 3

(ii) 3x² – 13x – 10 = 0; 5,(-2/3)

**Answer:**

**(i) x² – 5x + 6 = 0; 2, – 3**

**(ii) 3x² – 13x – 10 = 0; 5,(-2/3)**

**Question- 2.** **In each of the following, determine whether the given numbers are solutions of the given equation or not:**

**(i) x² – 3√3x + 6 = 0; √3, – 2√3**

**(ii) x² – √2x – 4 = 0, x = – √2, 2√2**

**Answer :**

**(i) x² – 3√3x + 6 = 0; √3, -2√3**

Substituting the value of x = √3

x^{2} – 3√3x + 6 = 0

(√3)^{2} – 3√3(√3) + 6 = 0

3 – 9 + 6 = 0

-9 + 9 =0

0 = 0

∴ √3 is the solution of the equation.

When, x = -2√3

x^{2} – 3√3x + 6 = 0

(-2√3)^{2} – 3√3(-2√3) + 6 = 0

4(3) +18 + 6 = 0

12 + 18 + 6 = 0

36 =0

∴ -2√3 is not the solution of the equation.

**(ii) x**^{2} – √2x – 4 = 0; x = -√2, 2√2

^{2}– √2x – 4 = 0; x = -√2, 2√2

Let us substitute the given values in the expression and check,

When, x = -√2

x^{2} – √2x – 4 = 0

(-√2)^{2} – √2(-√2) – 4 = 0

2 + 2 – 4 = 0

4 – 4 = 0

0 = 0

∴ -√2 is the solution of the equation.

When, x = 2√2

x^{2} – √2x – 4 = 0

(2√2)^{2} – √2(2√2) – 4 = 0

4(2) – 4 – 4 = 0

4 – 4 = 0

0 = 0

∴ 2√2 is the solution of the equation

**Question- 3.**

(i) If (-1/2) is a solution of the equation 3x² + 2kx – 3 = 0, find the value of k.

(ii) If 2/3 is a solution of the equation 7x² + kx – 3 = 0, find the value of k.

**Answer:**

**(i) **If –1/2 is a solution of the equation 3x² + 2kx – 3 = 0, find the value of k.

Let us substitute the given value x = -1/2 in the expression, we get

3x² + 2kx – 3 = 0

3(-1/2)^{2} + 2k(-1/2) – 3 = 0

3/4 – k – 3 = 0

¾ – 3 = k

By taking LCM

k = (3-12)/4

= -9/4

∴ Value of k = -9/4.

**(ii) **If 2/3 is a solution of the equation 7x² + kx – 3 = 0, find the value of k.

Let us substitute the given value x = 2/3 in the expression, we get

7x² + kx – 3 = 0

7(2/3)^{2} + k(2/3) – 3 = 0

7(4/9) + 2k/3 – 3 = 0

28/9 – 3 + 2k/3 = 0

2k/3 = 3 – 28/9

By taking LCM on the RHS

2k/3 = (27 – 28)/9

= -1/9

k = -1/9 × (3/2)

= -1/6

∴ Value of k = -1/6.

**Question -4.**

(i) If √2 is a root of the equation kx² + √2 – 4 = 0, find the value of k.

(ii) If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.

**Answer:**

**(i) kx² + √2 – 4 = 0, x = √2**

x = √2 is its solution

**(i) **If √2 is a root of the equation kx² + √2x – 4 = 0, find the value of k.

Let us substitute the given value x = √2 in the expression, we get

kx² + √2x – 4 = 0

k(√2)^{2} + √2(√2) – 4 = 0

2k + 2 – 4 = 0

2k – 2 = 0

k = 2/2

= 1

∴ Value of k = 1.

**(ii) **If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.

Let us substitute the given value x = a in the expression, we get

x² – (a + b)x + k = 0

a^{2} – (a + b)a + k = 0

a^{2} – a^{2} – ab + k = 0

-ab + k = 0

k = ab

∴ Value of k = ab.

**Question -5. ****If 2/3 and – 3 are the roots of the equation px² + 7x + q = 0, find the values of p and q.**

**Answer -5**

2/3 and – 3 are the roots of the equation px² + 7x + q = 0

Substituting the value of x = and – 3 respectively, we get

px² + 7x + q = 0

p(2/3)^{2} + 7(2/3) + q = 0

4p/9 + 14/3 + q = 0

By taking LCM

4p + 42 + 9q = 0

4p + 9q = – 42 … (1)

Now, substitute the value x = -3 in the expression, we get

px² + 7x + q = 0

p(-3)^{2} + 7(-3) + q = 0

9p + q – 21 = 0

9p + q = 21

q = 21 – 9p…. (2)

By substituting the value of q in equation (1), we get

4p + 9q = – 42

4p + 9(21 – 9p) = -42

4p + 189 – 81p = -42

189 – 77p = -42

189 + 42 = 77p

231 = 77p

p = 231/77

**p = 3**

Now, substitute the value of p in equation (2), we get

q = 21 – 9p

= 21 – 9(3)

= 21 – 27

**= -6**

— : End of ML Aggarwal Quadratic Equations Exe-5.1 Class 10 ICSE Maths Solutions :–

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