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RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5

September 1, 2021 by Fazal Leave a Comment

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) 2x2 – 3x + 5 = 0
(ii) 2x2 – 4x + 3 = 0
(iii) 2x2 + x – 1 = 0
(iv) x2 – 4x + 4 = 0
(v) 2x2 + 5x + 5 = 0
(vi) 3x2 – 2x + \(\frac { 1 }{ 3 }\) = 0
Solution
(i) Given equation 2x2 – 3x + 5 = 0
Comparing with ax2 + bx + c = 0
a = 2, b = – 3 and c = 5
Discriminant (D) = b2 – 4ac
= (-3)2 – 4 × 2 × 5
= 9 – 40
= -31 < 0
Thus given equation (RBSESolutions.com) has no real roots.
(ii) Given equation
2x2 – 4x + 3 =0
Comparing it with ax2 + bx + c = 0
a = 2, b = – 4 and c = 3
Discriminant(D) = b2 – 4ac
= (-4)2 – 4 × 2 × 3
= 16 – 24
= -8 < 0
Thus given equation has no real roots.
(iii) Given equation 2x2 + x – 1 = 0
Comparing it with ax2 + bx + c = 0
a = 2, b = 1 and c = – 1
Discriminant (D) = b2 – 4ac
= (1)2 – 4 × 2 × (-1)
= 1 + 8
= 9 > 0
Hence, roots of equation (RBSESolutions.com) are real and distinct
(iv) Given equation x2 – 4x + 4 = 0
Comparing it with quadratic equation ax2 + bx + c = 0
a = 1, b = -4 and c = 4
Discriminant (D) = b2 – 4ac
= (-4)2 – 4 × 1 × 4
= 16 – 16
= 0
Hence, roots of equation will be real and equal.
(v) Given equation 2x2 + 5x + 5 = 0
Comparing it with quadratic equation ax2 + 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 3x2 – 2x + \(\frac { 1 }{ 3 }\) = o
Comparing it by general quadratic equation
a = 3, b = -2, c = \(\frac { 1 }{ 3 }\)
Discriminant (D) = b2 – 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.

RBSE Solutions

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) x2 – 2(k + 1)x + k2 = 0
(iii) 2x2 + kx + 3 = 0
(iv) (k + 1)x2 – 2(k – 1) x + 1 = 0
(v) (k + 4)x2 + (k + 1) x + 1 = 0
(vi) kx2 – 5x + k = 0
Solution
(i) Given equation kx(x – 2) + 6 = 0
or kx2 – 2kx + 6 = 0
Comparing it with ax2 + bx + c = 0
a = k, b = -2k and c = 6
Discriminant D = b2 – 4ac
= (-2k)2 – 4 × k × 6
= 4k2 – 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
x2 – 2(k + 1)x + k2 = o
Comparing it with ax2 + bc + c = 0
a = 1, b = -2(k + 1) and c = k2
Discriminant D = b2 – 4ac
= {-2(k + 1)}2 – 4 × 1 × k2
= 4(k2 + 2k + 1) – 4k2
= 4k2 + 8k + 4 – 4k2
= 8k + 4
For equal roots D = 0
⇒ 8k + 4 = 0
⇒ 8k = -4
⇒ k = \(\frac { -1 }{ 2 }\)
Thus k = \(\frac { -1 }{ 2 }\)
(iii) Given equation
2x2 + kr + 3 = 0
Comparing with ax2 + bx + c = 0
a = 2, b = k and c = 3
Discriminant D = b2 – 4ac
= k2 – 4 × 2 × 3
= k2 – 24
for equal roots D = 0
⇒ k2 – 24 = 0
⇒ k2 = 24
⇒ k = ±√24 = ±2√6
for equal roots k = ± 2√6
(iv) Given (RBSESolutions.com) equation (k + 1)x2 – 2(k – 1)x + 1 = 0
Comparing it by ax2 + bx + c = 0
a = (k + 1), b = -2(k – 1) and c = 1
Discriminant, D = b2 – 4ac
= {-2 {k – 1)}2 – 4 × (k + 1) × 1
= 4{k2 + 1 – 2k) – 4 (k + 1)
= 4k2 + 4 – 8k – 4k – 4
= 4k2 – 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)x2 + (k + 1)x + 1 = 0
Comparing it with ax2 + bx + c = 0
a = k + 4, b = k + 1, c = 1
Discriminant (D) = b2 – 4ac
= (k + 1 )2 – 4 × (k + 4) × 1
= k2 + 2k + 1 – 4k – 16
= k2 – 2k – 15
= k2 – 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 kx2 – 5x + k = 0
Comparing it with ax2 + bx + c = 0
a = k, b = -5. c = k
Discriminant, (D) = b2 – 4ac
= (-5)2 – 4 × k × k
= 25 – 4k2
For equal roots D = 0
RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5 1

RBSE Solutions

Question 3.
Find value of k for which (RBSESolutions.com) following quadratic equations have real and fractional roots.
(i) kx2 + 2x + 1 = 0
(ii) kx2 + 6x + 1 = 0
(iii) x2 – kx + 9 = 0
Solution
(i) Given roots kx2 + 2x + 1 = 0
Comparing it by general quadratic equation
ax2 + 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 kx2 + 6x + 1 = 0
Comparing it by general (RBSESolutions.com) quadratic equation ax2 + bx + c – 0,
a = k, b = 6, c = 1
Discriminant (D) = b2 – 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 x2 – kx + 9 = 0
Comparing it by general quadratic equation
ax2 + bx + c = 0
a = 1, b = -k, c = 9
Discriminant (D) = b2 – 4ac
= (-k)2 – 4 × 1 × 9
= k2 – 36
If D > 0, then roots of (RBSESolutions.com) equation will be real and fractional.
D > 0
⇒ k2 – 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 x2 + 5kx + 16 = 0 has no real roots.
Solution
Given equation x2 + 5kx + 16 = 0
Comparing it by general quadratic equation ax2 + bx + c = 0
a = 1, b = 5k, c = 16
Discriminant (D) = b2 – 4ac
= (5k)2 – 4 × 1 × 16
= 25k2 – 64
If Discriminant (D) < 0, then roots will not be real.
i.e. D < 0
⇒ 25k2 – 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)x2 + (c – a)x + (a – b) = 0 are real and equal, then prove that 2b = a + c
Solution
Given equation is (b – c)x2 + (c – a)x + (a – b) = 0
Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b)
Discriminant (D) = b2 – 4ac
= (c – a)2 – 4(b – c)(a – b)
= (c2 + a2 – 2ac) – 4(ab – ac – b2 + bc)
= c2 + a2 – 2ac – 4ab + 4ac + 4b2 – 4bc
= c2 + a2 + 2ac – 4ab – 4bc + 4b2
= (a + c)2 – 4 b(a + c) + 4b2
= (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.

RBSE Solutions

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.

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