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

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.

The Polynomial Root Calculator can find multiple roots to polynomial equations with just one click. To open the tool, click on it.

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

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

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 \(\frac { -8 }{ 5 }\)

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.

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