RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.5.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomials |

E×ercise |
Exercise 3.5 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.5

RBSE Class 10 Maths Exercise 3.5 Solutions Question 1.

Find the nature of roots of following (RBSESolutions.com) quadratic equations.

(i) 2x^{2} – 3x + 5 = 0

(ii) 2x^{2} – 4x + 3 = 0

(iii) 2x^{2} + x – 1 = 0

(iv) x^{2} – 4x + 4 = 0

(v) 2x^{2} + 5x + 5 = 0

(vi) 3x^{2} – 2x + \(\frac { 1 }{ 3 }\) = 0

Solution

(i) Given equation 2x^{2} – 3x + 5 = 0

Comparing with ax^{2} + bx + c = 0

a = 2, b = – 3 and c = 5

Discriminant (D) = b^{2} – 4ac

= (-3)^{2} – 4 × 2 × 5

= 9 – 40

= -31 < 0

Thus given equation (RBSESolutions.com) has no real roots.

(ii) Given equation

2x^{2} – 4x + 3 =0

Comparing it with ax^{2} + bx + c = 0

a = 2, b = – 4 and c = 3

Discriminant(D) = b^{2} – 4ac

= (-4)^{2} – 4 × 2 × 3

= 16 – 24

= -8 < 0

Thus given equation has no real roots.

(iii) Given equation 2x^{2} + x – 1 = 0

Comparing it with ax^{2} + bx + c = 0

a = 2, b = 1 and c = – 1

Discriminant (D) = b^{2} – 4ac

= (1)^{2} – 4 × 2 × (-1)

= 1 + 8

= 9 > 0

Hence, roots of equation (RBSESolutions.com) are real and distinct

(iv) Given equation x^{2} – 4x + 4 = 0

Comparing it with quadratic equation ax^{2} + bx + c = 0

a = 1, b = -4 and c = 4

Discriminant (D) = b^{2} – 4ac

= (-4)^{2} – 4 × 1 × 4

= 16 – 16

= 0

Hence, roots of equation will be real and equal.

(v) Given equation 2x^{2} + 5x + 5 = 0

Comparing it with quadratic equation ax^{2} + bx + c = 0

a = 2, b = 5 and c = 5

Discriminant (D) = (5)^{2} – 4 × 2 × 5

= 25 – 40

= -15 < 0

Thus, roots of equation are not real.

(vi) Given equation 3x^{2} – 2x + \(\frac { 1 }{ 3 }\) = o

Comparing it by general quadratic equation

a = 3, b = -2, c = \(\frac { 1 }{ 3 }\)

Discriminant (D) = b^{2} – 4ac

= (-2)^{2} – 4 × 3 × \(\frac { 1 }{ 3 }\)

= 4 – 4

= 0

Hence, root of given equation are real and equal.

RBSE Solutions For Class 10 Maths Chapter 3 Question 2.

Find value of k, for which (RBSESolutions.com) following quadratic equations have real and equal roots.

(i) kx(x – 2) + 6 = 0

(ii) x^{2} – 2(k + 1)x + k^{2} = 0

(iii) 2x^{2} + kx + 3 = 0

(iv) (k + 1)x^{2} – 2(k – 1) x + 1 = 0

(v) (k + 4)x^{2} + (k + 1) x + 1 = 0

(vi) kx^{2} – 5x + k = 0

Solution

(i) Given equation kx(x – 2) + 6 = 0

or kx^{2} – 2kx + 6 = 0

Comparing it with ax^{2} + bx + c = 0

a = k, b = -2k and c = 6

Discriminant D = b^{2} – 4ac

= (-2k)^{2} – 4 × k × 6

= 4k^{2} – 24k

= 4k(k – 6)

For equal roots D = 0

4k(k – 6) = 0

⇒ k = 0 or k – 6 = 0

⇒ k = 0 or k = 6

For equal roots k = 6 because k = 0 is not possible.

(ii) Given (RBSESolutions.com) equation

x^{2} – 2(k + 1)x + k^{2} = o

Comparing it with ax^{2} + bc + c = 0

a = 1, b = -2(k + 1) and c = k^{2}

Discriminant D = b^{2} – 4ac

= {-2(k + 1)}^{2} – 4 × 1 × k^{2}

= 4(k^{2} + 2k + 1) – 4k^{2}

= 4k^{2} + 8k + 4 – 4k^{2}

= 8k + 4

For equal roots D = 0

⇒ 8k + 4 = 0

⇒ 8k = -4

⇒ k = \(\frac { -1 }{ 2 }\)

Thus k = \(\frac { -1 }{ 2 }\)

(iii) Given equation

2x^{2} + kr + 3 = 0

Comparing with ax^{2} + bx + c = 0

a = 2, b = k and c = 3

Discriminant D = b^{2} – 4ac

= k^{2} – 4 × 2 × 3

= k^{2} – 24

for equal roots D = 0

⇒ k^{2} – 24 = 0

⇒ k^{2} = 24

⇒ k = ±√24 = ±2√6

for equal roots k = ± 2√6

(iv) Given (RBSESolutions.com) equation (k + 1)x^{2} – 2(k – 1)x + 1 = 0

Comparing it by ax^{2} + bx + c = 0

a = (k + 1), b = -2(k – 1) and c = 1

Discriminant, D = b^{2} – 4ac

= {-2 {k – 1)}^{2} – 4 × (k + 1) × 1

= 4{k^{2} + 1 – 2k) – 4 (k + 1)

= 4k^{2} + 4 – 8k – 4k – 4

= 4k^{2} – 12k

= 4k(k – 3)

for equal roots, D = 0

⇒ 4k(k – 3) = 0

⇒ k(k – 3) =0

⇒ k = 0 or k = 3

for equal roots k = 3, since k = 0

(v) Given equation is (k + 4)x^{2} + (k + 1)x + 1 = 0

Comparing it with ax^{2} + bx + c = 0

a = k + 4, b = k + 1, c = 1

Discriminant (D) = b^{2} – 4ac

= (k + 1 )^{2} – 4 × (k + 4) × 1

= k^{2} + 2k + 1 – 4k – 16

= k^{2} – 2k – 15

= k^{2} – 5k + 3k – 15

= k(k – 5) + 3(k – 5)

= (k – 5) (k + 3)

For equal roots, D = 0

⇒ (k – 5)(k + 3) = 0

⇒ k – 5 = 0 or k + 3 = 0

⇒ k = 5 or k = -3

(vi) Given (RBSESolutions.com) equation kx^{2} – 5x + k = 0

Comparing it with ax^{2} + bx + c = 0

a = k, b = -5. c = k

Discriminant, (D) = b^{2} – 4ac

= (-5)^{2} – 4 × k × k

= 25 – 4k^{2}

For equal roots D = 0

RBSE Solutions For Class 10 Maths Chapter 3.5 Question 3.

Find value of k for which (RBSESolutions.com) following quadratic equations have real and fractional roots.

(i) kx^{2} + 2x + 1 = 0

(ii) kx^{2} + 6x + 1 = 0

(iii) x^{2} – kx + 9 = 0

Solution

(i) Given roots kx^{2} + 2x + 1 = 0

Comparing it by general quadratic equation

ax^{2} + bx + c = 0

a = k, b = 2, c = 1

Discriminant (D) = b2 – 4ac

= (2)^{2} – 4 × k × 1

= 4 – 4 k

= 4(1 – k)

For real and fractional roots

Discriminant (D) > 0

⇒ 4(1 -k) > 0

⇒ 1 – k > 0

⇒ k < 1

Thus, k will be less than.

(ii) Given equation kx^{2} + 6x + 1 = 0

Comparing it by general (RBSESolutions.com) quadratic equation ax^{2} + bx + c – 0,

a = k, b = 6, c = 1

Discriminant (D) = b^{2} – 4ac

= (6)^{2} – 4 × k × 1

= 36 – 4k

= 4(9 – k)

If discriminant (D) > 0 then roots of equation will be real and fractional.

i.e., D > 0

⇒ 4(9 – k) > 0

⇒ 9 – k > 0

⇒ k > 9

Thus value of k will be greater than 9

(iii) Given equation x^{2} – kx + 9 = 0

Comparing it by general quadratic equation

ax^{2} + bx + c = 0

a = 1, b = -k, c = 9

Discriminant (D) = b^{2} – 4ac

= (-k)^{2} – 4 × 1 × 9

= k^{2} – 36

If D > 0, then roots of (RBSESolutions.com) equation will be real and fractional.

D > 0

⇒ k^{2} – 36 > 0

⇒ (k – 6)(k + 6) > 0

⇒ k – 6 > 0 or k + 6 > 0

⇒ k > 6 or k < -6

Thus, value of k will be greater than 6 or less than -6

Ex 3.5 Class 10 RBSE Question 4.

Find value of k, for which equation x^{2} + 5kx + 16 = 0 has no real roots.

Solution

Given equation x^{2} + 5kx + 16 = 0

Comparing it by general quadratic equation ax^{2} + bx + c = 0

a = 1, b = 5k, c = 16

Discriminant (D) = b^{2} – 4ac

= (5k)^{2} – 4 × 1 × 16

= 25k^{2} – 64

If Discriminant (D) < 0, then roots will not be real.

i.e. D < 0

⇒ 25k^{2} – 64 < 0

⇒ (5k – 8)(5k + 8) < 0

⇒ 5k – 8 < 0 or 5k + 8 < 0

⇒ k < \(\frac { 8 }{ 5 }\) or k > \(\frac { -8 }{ 5 }\)

Hence, value of k will be smaller than \(\frac { 8 }{ 5 }\) or greater than \(\frac { -8 }{ 5 }\)

Exercise 3.5 Class 10 RBSE Question 5.

If roots of quadratic (RBSESolutions.com) equation (b – c)x^{2} + (c – a)x + (a – b) = 0 are real and equal, then prove that 2b = a + c

Solution

Given equation is (b – c)x^{2} + (c – a)x + (a – b) = 0

Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b)

Discriminant (D) = b^{2} – 4ac

= (c – a)^{2} – 4(b – c)(a – b)

= (c^{2} + a^{2} – 2ac) – 4(ab – ac – b^{2} + bc)

= c^{2} + a^{2} – 2ac – 4ab + 4ac + 4b^{2} – 4bc

= c^{2} + a^{2} + 2ac – 4ab – 4bc + 4b^{2}

= (a + c)^{2} – 4 b(a + c) + 4b^{2}

= (a + c)^{2} – 2 × (a + c) × 2b + (2b)^{2}

= [(a + c) – 2b]^{2}

Roots of equation (RBSESolutions.com) are real and equal.

D = 0

⇒ [(a + c) – 2b]^{2} = 0

⇒ (a + c) – 2b = 0

⇒ a + c = 2b

Hence Proved.

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.5, drop a comment below and we will get back to you at the earliest.