WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.5
Question 1. Let us write the nature of two roots of the following quadratic equations:
1. 2x2 + 7x+3=0
Solution: Here, Discriminant = \(b^2-4 a c\)
= \((7)^2-4.2 \cdot 3\)
= 49 – 24 = 25 > 0
∴ Roots are real & unequal.
2. 3x2-2√6x+2=0
Solution: Discriminant = \(b^2-4 a c\)
= \((-2 \sqrt{6})^2-4.3 .2\)
= 24 – 24 – 0
∴ Roots are real & equal.
Read and Learn More WBBSE Solutions For Class 10 Maths
3. 2x2– 7x+9=0
Solution: Discriminant = b2 – 4ac
= (-7)2 – 4.2.9
= 49 – 72
= -23 < 0
∴ the equation has no real roots.
| Chapter Name | Solutions | ||||
| Chapter 1 Quadratic Equations in One Variable | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | Exercise 5 |
| Chapter 2 Simple Interest | Exercise 1 | Exercise 2 | |||
| Chapter 3 Theorems Related to Circle | Exercise 1 | Exercise 2 | |||
| Chapter 4 Rectangular Parallelopiped or Cuboid | Exercise 1 | Exercise 2 | |||
| Chapter 5 Ratio and Proportion | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease | Exercise 1 | Exercise 2 | |||
| Chapter 7 Theorems Related To Angles In A Circle | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | Exercise 5 |
| Chapter 8 Right Circular Cylinder | Exercise 1 | Exercise 2 | |||
| Chapter 9 Quadratic Surd | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 10 Theorems Related To Cyclic Quadrilateral | Exercise 1 | ||||
| Chapter 11 Construction Of Circumcircle and Incircle Of A Traingle | Exercise 1 | Exercise 2 | |||
| Chapter 12 Sphere | Exercise 1 | Exercise 2 | |||
| Chapter 13 Variation | Exercise 1 | Exercise 2 | |||
| Chapter 14 Partnership Business | Exercise 1 | ||||
| Chapter 15 Theorems Related To Tangent To A Circle | Exercise 1 | Exercise 2 | |||
| Chapter 16 Right Circular Cone | Exercise 1 | ||||
| Chapter 17 Construction Of Tangent To A Circle | Exercise 1 | ||||
| Chapter 18 Similarity | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | |
| Chapter 19 Problems Related To Different Solid Objects | Exercise 1 | ||||
| Chapter 20 Trigonometry: Concept of Measurement Of Angle | Exercise 1 | ||||
| Chapter 21 Construction: Determination Of Mean Proportional | Exercise 1 | ||||
| Chapter 22 Pythagoras Theorem | Exercise 1 | ||||
| Chapter 23 Trigonometric Ratios And Trigonometric Identities | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 24 Trigonometric Ratios Of Complementary Angle | Exercise 1 | ||||
| Chapter 25 Application Of Trigonometric Rations: Heights & Distances | Exercise 1 | ||||
| Chapter 26 Statistics: Mean, Median, Ogive, Mode | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 |
Question 2. By calculating, let us write the value/values of k for which each of the following quadratic equations has real and equal roots-
1. 3x2-5x+2k = 0
Solution: As the roots of the equation are real & equal.
∴ b2 – 4ac = 0
i.e., (-5)2 – 4.3.2k = 0
or, 25 – 24k = 0
∴ 24k = 25
∴ k = \(\frac{25}{24}\)
2. 9x2-24x + k = 0
Solution: As the roots of the equation are real & equal.
∴ Discriminant = b2 – 4ac = 0
i.e., (-24)2 – 4.9.k = 0
or, 576 – 36k = 0
∴ 36k = 576
∴ k = \(\frac{576}{36}\) = 16
3. 2x2 + 3x + k=0
Solution: As the roots of the equation are real & equal.
∴ Discriminant = b2 – 4ac = 0
i.e., (3)2 – 4.2.k = 0
or, 9 – 8k = 0
∴ 8k = 9
∴ k = \(\frac{9}{8}\)
4. x2-2 (5+ 2k) x + 3 (7+10k) = 0
Solution: As the roots of the equation are real & equal.
∴ Discriminant, b2 – 4ac = 0
i.e., {-2(5 + 2k)}2 – 4.1.3(7 + 10k) = 0
or, 4(25 + 20k + 4k2) – 12(7 + 10k) = 0
or, 100 + 80k + 16k2 – 84 – 120k = 0
or, 16k2 – 40k + 16 = 0
or, 8(2k2 – 5k + 2) = 0
or, 2k2 – 4k – k + 2 = 0
or, 2k(5k – 2) – 1(k – 2) = 0
or, (k – 2)(2k – 1) = 0
Either k – 2 = 0
∴ k = 2
2k – 1 = 0
∴ k = 1/2
5. (3k+ 1)x2 + 2 (k+1)x + k = 0
Solution: As the roots of the equation are real & equal.
∴ Discriminant b2 – 4ac = 0
i.e., {2(k + 1)}2 – 4.(3k + 1).k = 0
or, 4(k2 + 2k + 1) – 12k2 – 4k = 0
or, 4k2 + 8k + 4 – 12k2 – 4k = 0
or, -8k2 + 4k + 4 = 0
or, -4(2k2 – k – 1) = 0
or, 2k2 – k – 1 = 0
or, 2k2 – 2k + k – 1 = 0
or, 2k(k – 1) + 1(k – 1) = 0
or, (k – 1)(2k + 1) = 0
Either k – 1 = 0
∴ k = 1
2k + 1 = 0
∴ k = -1/2
Question 3. Let us form the quadratic equations from two roots given below-
1. 4, 2
Solution: When roots are 4, 2:1
The required equation: x2 (Sum of the roots) x + product of roots = 0 i.e., x2 – (4+2)x + 4.2 = 0
or, x2-6x+8=0
2.-4, -3
Solution: When roots are -4, -3:
The required equation:- x2 – (Sum of the roots) x + product of the roots = 0
i.e. ; x2(43) x + (-4) (-3)=0
or, x2 + 7x+12=0
3. -4, 3
Solution: When roots are -4, 3:
The required equation: x2 – (Sum of the roots) x + product of the roots = 0
i.e., x2-(-4+3) x + (-4).3 = 0
or, x2 + 1x-12=0
or, x2+x-12=0
4. 5, -3
Solution: When roots are 5, -3:
The required equation: x2 – (Sum of the roots) x + product of the roots = 0
i.e., x2 (5-3) x + 5 (-3)=0
or, x2-2x-15=0
Question 4. If two roots of the quadratic equation (a2+ b2)x2 -2 (ac + bd)x + (c2 + d2) = 0 are equal, then let us prove that, a/c = b/d
Solution: (a2 + b2)x2 – 2(ac + bd)x + (c2 + d2) = 0.
As the roots of the equation are equal.
∴ Discriminant b2 – 4ac = 0
i.e., {-2(ac + bd)}2 – 4(a2 + b2)(c2 + d2) = 0
or, 4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2 + b2d2) = 0
or, 4a2c2 + 4b2d2 + 8abcd – 4a2c2 – 4a2d2 – 4b2c2 – 4b2d2 = 0
or, 8abcd – 4a2d2 – 4b2c2 = 0
or, -4(a2d2 – 2abdc + b2c2) = 0
or, (ad – bc)2 = 0
∴ ad = bc
∴ \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}\) proved.
Question 5. Let us prove that the quadratic equation 2(a2 + b2)x2 + 2 (a + b) x + 1 = 0 has no real root if a b.
Solution: 2(a2 + b2)x2 + 2(a + b)x + 1 = 0
As the equation has no real roots.
∴ Discriminant b2 – 4ac < 0
i.e., {2(a + b)}2 – 4.2(a2 + b2).1 < 0
or, 4a2 + 4b2 + 8ab – 8a2 – 8b2 < 0
or, -4a2 – 4b2 + 8ab < 0
or, -4(a2 + b2 – 2ab) < 0
or, a2 + b2 – 2ab > 0
or, (a – b)2 > 0
∴ a – b > 0
i.e., a ≠ b Proved.
Question 6. If two roots of the quadratic equation 5x2+2x-3= 0 are x and ẞ, then let us determine the value of :
Solution: As x and are the roots of the equation 5x2 + 2x – 3 = 0
∵ \(\alpha+\beta=-\frac{2}{5} \text { and } \alpha \beta=-\frac{3}{5}\)
1. α2+ β2
Solution: \((\alpha+\beta)^2-2 \alpha \beta\)
= \(\left(-\frac{2}{5}\right)^{2^{\prime}}-2\left(-\frac{3}{5}\right)^2\)
= \(\frac{4}{25}+\frac{6}{5}=\frac{4+30}{25}=\frac{34}{25}\)
(2) α3 +ẞ3
Solution: \((\alpha+\beta)^3-3 \propto \beta(\alpha+\beta)\)
= \(\left(-\frac{2}{5}\right)^3-3\left(-\frac{3}{5}\right)\left(-\frac{2}{5}\right)\)
= \(-\frac{8}{125}-\frac{18}{25}=\frac{-8-90}{125}=\frac{-98}{125}\)
(3) α2 /B + B2/ α
Solution: \(\frac{\alpha^3+\beta^3}{\alpha \beta}=\left(\frac{-98}{125}\right) \div\left(\frac{-3}{5}\right)\)
= \(\frac{-98}{125} \times\left(\frac{-5}{3}\right)=+\frac{98}{75}\)
Question 7. If one root of the equation ax2 + bx + c = 0 is twice of the other, then let us show that 2b2 = 9ac.
Solution: ax2 + bx + c = 0
Let one root is β, then the other root = 2α.
∴ \(\alpha+2 \alpha=-\frac{b}{a}\)
or, \(3 \propto=-\frac{b}{a}\)
or, \(\alpha=-\frac{b}{3 a}\)
Again, \(\alpha .2 \alpha=\frac{\mathrm{c}}{\mathrm{a}}\)
or, \(2 \propto^2=\frac{c}{a}\)
∴ \(2\left(-\frac{b}{3 a}\right)^2=\frac{c}{a}\)
\(\frac{2 b^2}{9 a^2}=\frac{c}{a}\)∴ 2b2 = 9ac proved.
Question 8. Let us form the equation whose roots are reciprocals to the roots of equation x2 + px + 1 = 0.
Solution: x2 + px + 1 = 0
Let the roots of equation be α and β.
∴ \(\alpha+\beta=\frac{-p}{1}=-p\)
& \(\alpha \beta=\frac{1}{1}=1\)
Now, the roots of the new equation are \(\frac{1}{\alpha} \text { and } \frac{1}{\beta}.\)
∴ Some of the roots = \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-p}{1}=-p\)
Product of the roots = \(\frac{1}{\alpha} \cdot \frac{1}{\beta}=\frac{1}{\alpha \beta}=\frac{1}{1}=1\)
∴ The required equation is x2 – (sum of the roots)x + product of the roots = 0
or, x2 – (-p)x + 1 = 0 i.e., x2 + px + 1 = 0
Chapter 1 Quadratic Equations In One Variable Exercise 1.5 Multiple Choice Questions
Question 1. The sum of the two roots of the equation x2-6x + 2 = 0 is
1. 2
2. -2
3. 6
4. -6
Solution: Sum of the roots = – (-6)/1= 6………………(c)
Question 2. If the product of two roots of the equation x2 – 3x + k = 10 is -2, then the value of k is
1. -2
2. -8
3. 8
4 12
Solution:
Given
x2 – 3x + (k-10) = 0
Product of the roots = k -’10
K-10=-2 k10-28 —–(c)
Question 3. If two roots of the equation ax2 + bx + c = 0 (a + 0) are real and unequal, then b2 -4ac will be
1. > 0
2. = 0
3. < 0
4. None of these
Solution: The roots of the equation ax2 + bx + c = 0 (a + 0)
are real & unequal if b2-4ac >0—– (a).
Product of the roots = k -’10
k-10=-2 k10-28 —–(c)
Question 4. If two roots of the equation ax2 + bx + c = 0 (a + 0) are real and unequal, then b2 -4ac will be
1. > 0
2. = 0
3. < 0
4. None of these
Solution: The roots of the equation ax2 + bx + c = 0 (a + 0)
are real & unequal if b2-4ac >0—– (a).
Question 5. If two roots of the equation ax2 + bx + c = 0 (a + 0) be equal, then
1. c=-b/2a
2. c=b/2a
3. c= -b2/4a
4. c=b2/4a
Solution: The roots of the equation ax2 + bx + c = 0 will be equal if b2 = 4ac.
or, c =b2/4a
Question 6. If two roots of the equation 3x2 + 8x+2=0 be∞ and ẞ, then the value of (1/∞+1/ ẞ) is
1. -3/8
2. 2/3
3. -4
4. 4
Solution: The roots of the equation 3x2 + 8x + 2 = 0
are α and β then \(\frac{1}{\alpha}+\frac{1}{\beta}\)
= \(\frac{\alpha+\beta}{\alpha \beta}\)
= \(\frac{-\frac{8}{3}}{\frac{2}{3}}\)
= –\(\frac{8}{2}\)
= -4
Chapter 1 Quadratic Equations In One Variable Exercise 1.5 True or False:
Question 1. The roots of equation x2 + x + 1 = 0 are real.
Answer: False, as here b2-4ac1-4.1.1-4 <1.
Question 2. The roots of the equation x2 – x + 2 are not real.
Answer: True, as here b2-4ac = (-1)2-4.1.2-1-8-7 <1.
Chapter 1 Quadratic Equations In One Variable Exercise 1.5 Fill in the blanks:
Question 1. The ratio of the sum and the product of two roots of equation 7x2 – 12x + 18 = 0 ——————-
Solution: Sum of the roots = \(-\frac{-(12)}{7}=\frac{12}{7}\)
and product of the roots = \(\frac{18}{7}\)
∴ the ratio of sum & product.
= \(\frac{12}{7}: \frac{18}{7}\) = 12 : 18 = 2 : 3.
Question 2. If two roots of the equation ax2 + bx + c = 0 (a + 0) are reciprocal to each other, then c =—————–
Solution: As the roots of this equation are reciprocal
∴ \(\propto \frac{1}{\alpha}=\frac{\mathrm{c}}{\mathrm{a}} \text { or } \frac{\mathrm{c}}{\mathrm{a}}=1\)
∴ c = a
∴ a + c = 0
Question 3. If two roots of the equation ax2 + bx + c = 0 (a + 0) are reciprocal to each other and opposite (negative).
Solution: \(X \times\left(-\frac{2}{x}\right)=\frac{c}{a}\)
∴ \(\frac{c}{a}\) = -1
or, c = -9
∴ a + c = 0
Chapter 1 Quadratic Equations In One Variable Exercise 1.5 Short Answers
Question 1. Let us write the quadratic equation if the sum of its roots is 14 and the product of them is 24.
Solution: The sum of the roots. 14 & the product of the roots = 24.
The required equation is x2 – 14x + 24 = 0.
Question 2. If the sum and the product of two roots of the equation kx2 + 2x + 3k = 0 (k = 0) áre equal, let us write the value of k.
Solution: The sum of the roots = –\(\frac{2}{k}\)
and the product of roots = \(\frac{3 k}{k}\) = 3.
∴ –\(\frac{2}{k}\)
or, 3k = -2
∴ k = –\(\frac{2}{3}\)
Question 3. If x and ẞ be the two roots of the equation x2 – 22x + 105= 0, let us write the value of (x – B).
Solution: As α and β are the roots of the equation x2 – 22x + 105 = 0
∴ \(\alpha+\beta=-\left(\frac{-22}{1}\right)=22 \text { and } \alpha \beta=105\)
\((\alpha-\beta)^2=(\alpha+\beta)^2-4 \propto \beta=(22)^2-4 \times 105=484-420=64\)∴ \((\alpha-\beta)=\sqrt{64}= \pm 8\)
Question 4. If the sum of two roots of the equation x2-x= k (2x-1) is zero, let us write the value of k.
Solution: Given x2 – x = k(2x – 1)
or, x2 – x – 2kx + k = 0
or, x2 – (1 + 2k)x + k = 0
as the sum of the roots = 0
∴ 1 + 2k = 0
∴ k = –\(\frac{1}{2}\)
Question 5. If one of the roots of the two equations x2 + bx+12= 0 and x2 + bx + q = 0 is 2, let us write the value of q.
Solution: As 2 is the common root of the equation x2 + Dx + 12 = 0
i.e., (2)2 + b x 2 + 12 = 0
or, 2b = -16
∴ b – 8
Again, 2 is the root of the 2nd equation x2 + bx + q = 0
∴ (2)2 + (-8) x 2 + q = 0
or, 4 – 16 + q = 0
∴ q = 12.