WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.2
Question 1: In each of the following cases, let us justify & write whether the given values are the roots of the given quadratic equation:
1. x2+x+1=0
Solution: f(X) = x2+x+1=0
When x = 1 ;
(1)2+1+1=1+1+1 = 3
Again when x = -1 ;
(-1)2+1(-1)+1=1-1+1 = 1
(1) & (-1) are not the roots of the given equation
2. 8x2+7x=0
Solution: f(X) = 8x2+7x=0
When x = 0
8(0) +7(0) = 0
When x = -2
8(-2)2+7(-2) = 32-14= 18
0 & (-2) are not the roots of the given equation.
| 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 |
3. x +1/x =13/6
Solution: When x = \(\frac{5}{6}\)
∴ \(\frac{5}{6}+\frac{1}{5 / 6}=\frac{5}{6}+\frac{6}{5}=\frac{25+36}{30}=\frac{61}{30}\)
∴ \(\frac{4}{3}+\frac{1}{4 / 3}=\frac{4}{3}+\frac{3}{4}=\frac{16+9}{30}=\frac{25}{12}\)
And when x = \(\frac{4}{3}\)
5/6 & 4/3 are not the roots of the given equation.
4. x2– √3x-6=0
Solution: f(X) = x2– √3x-6=0
When x = – √3
(√3)2-√3(-√3)-6=3+3-6 = 0
& When x = 2√3,
(2√3)2-√3 (2√3)-6=12-6-6 = 0
(√3) & (2√3). are the roots of the given equation.
Question 2:
1. Let us calculate and write the value of k for which 2/3 will be a root the quadratic equation 7x2 + kx – 3=0.
Solution: \(7 x^2+k x-3=0\)
As \(\frac{2}{3}\) is one root of the equation \(7 x^2+k x-3=0\)
∴ \(7\left(\frac{2}{3}\right)^2+k \cdot \frac{2}{3}-3=0\)
⇒ \(7 \cdot \frac{4}{9}+\frac{k \cdot 2}{3}-3=0\)
⇒ \(\frac{28}{9}+\frac{2 k}{3}-3=0\)
⇒ \(\frac{2 k}{3}=3-\frac{28}{9}\)
⇒ \(\frac{2 k}{3}=\frac{-1}{9}\)
∴ \(k=\frac{-3}{9 \times 2}=-\frac{1}{6}\)
2. Let us calculate and write the value of k for which -a will be a root of the quadratic equation x2 + 3ax + k = 0.
Solution:
Given
x2 + 3ax + k = 0
As (-a) is the root of the equation
x2 + 3ax + k = 0
⇒ (-a)2 + 3a(-a) + k = 0
⇒ +a2-3a2 + k = 0
⇒ -2a2+ k = 0
k = 2a2 Solution
The value of k = 2a2
3. If 2/3 and -3 are the two roots of the quadratic equation ax2 + 7x + b = 0, then Let me calculate the values of a and b.
Solution: When x = \(\frac{2}{3}\)
∴ \(a\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+b=0\)
⇒ \(\frac{4 a}{9}+\frac{14}{3}+b=0\)
⇒ \(\frac{4 a+42+9 b}{9}=0\)
4a + 9b = -42
When x = -3
∴ \(a(-3)^2+7(-3)+b=0\)
9a – 21 + b = 0
9a + b = 21
Now, solving 4a + 9b = -42 & 9a + b = 21
we get, a = 3 & b = -6.
Question 3: Let us solve
1. 3y2-20 = 160 – 2y2
Solution: \(3 y^2-20=160-2 y^2\)
⇒ \(3 y^2+2 y^2=160+20\)
⇒ \(5 y^2=180\)
⇒ \(5 y^2-180=0\)
⇒ \(5\left(y^2-36\right)=0\)
⇒ \(y^2-36=0\)
⇒ \((y)^2-(6)^2=0\)
∴ (y + 6)(y – 6) = 0
∴ When y + 6 = 0 & when y – 6 = 0
2. (2x+1)2 + (x + 1)2 = 6x + 47
Solution: \((2 x+1)^2+(x+1)^2=6 x+47\)
⇒ \(4 x^2+4 x+1+x^2+2 x+1-6 x-47=0\)
⇒ \(5 x^2-45=0\)
⇒ \(5\left(x^2-9\right)=0\)
⇒ \((x)^2-(3)^2=0\)
⇒ (x + 3)(x – 3) = 0
∴ Either x + 3 = 0
x = -3
Or, x – 3 = 0
∴ x = -3
3. (x-7) (x-9) = 195
Solution: (x – 7)(x – 9) = 195
⇒ \(x^2-7 x-9 x+63-195=0\)
⇒ \(x^2-16 x-132=0\)
⇒ \(x^2-22 x+6 x-132=0\)
⇒ x(x – 22) + 6(x – 22) = 0
⇒ (x – 22)(x + 6) = 0
∴ Either x – 22 = 0
x = 22
or, x + 6 = 0
∴ x = -6.
4. 3x-24/x=x/3, x ≠ 0
Solution: \(\frac{3 x^2-24}{x}=\frac{x}{3}\)
⇒ \(\left(3 x^2-24\right) \times 3=x^2\)
⇒ \(9 x^2-72-x^2=0\)
⇒ \(8 x^2-72=0\)
⇒ \(8\left(x^2-9\right)=0\)
⇒ \(\left(x^2-9\right)=0\)
⇒ \((x)^2-(3)^2=0\)
⇒ (x + 3)(x – 3) = 0
Either x + 3 = 0
∴ x = -3
or, x – 3 = 0
∴ x = 3.
5. \(\frac{x-2}{x+2}+\frac{6(x-2)}{x-6}=1\), x ≠ -2 , 6
Solution: \(\frac{x-2}{x+2}+\frac{6(x-2)}{x-6}=1\)
⇒ \(\frac{(x-2)(x-6)+6(x-2)(x+2)}{(x+2)(x-6)}=1\)
⇒ \(x^2-8 x+12+6\left(x^2-4\right)=(x+2)(x-6)\)
⇒ \(7 x^2-8 x-12=x^2+2 x-6 x-12\)
⇒ \(7 x^2-8 x-12-x^2+4 x+12=0\)
⇒ \(6 x^2-4 x=0\)
⇒ 2x(3x – 2) = 0
Either 2x = 0
∴ x = 0
Or, 3x – 2 = 0
∴ x = 2/3
6. \(\frac{x+5-(x-3)}{(x-3)(x+5)}=\frac{1}{6}\), x ≠ 3,-5
Solution: \(\frac{x+5-(x-3)}{(x-3)(x+5)}=\frac{1}{6}\)
⇒ \(\frac{x+5-x+3}{x^2+5 x-3 x-15}=\frac{1}{6}\)
⇒ \(x^2+2 x-15=48\)
⇒ \(x^2+2 x-63=0\)
⇒ \(x^2+9 x-7 x-63=0\)
⇒ x(x + 9) – 7(x + 9) = 0
⇒ (x + 9)(x – 7) = 0
∴ Either x + 9 = 0
∴ x = -9
or, x – 7 = 0
∴ x = 7.
7. \(\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{1}{12}\), x ≠ 0 , -1
Solution: \(\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{1}{12}\)
⇒ \(\frac{x^2+(x+1)^2}{x(x+1)}=\frac{25}{12}\)
⇒ \(\left(x^2+x^2+2 x+1\right) \times 12=25\left(x^2+x\right)\)
⇒ \(24 x^2+24 x+12=25 x^2+25 x\)
⇒ \(24 x^2-25 x^2+24 x-25 x+12=0\)
⇒ \(-x^2-x+12=0\)
⇒ \(x^2+x-12=0\)
⇒ (x + 4).(x – 3) = 0
Either x + 4 = 0
∴ x = -4
or, x – 3 = 0
∴ x = 3.
8. \(\frac{a x+b}{a+b x}=\frac{c x+d}{c+d x}\)[x≠ b, c≠ d] , x≠ a/b , -c/d
Solution: \(\frac{a x+b}{a+b x}=\frac{c x+d}{c+d x}\)
⇒ (ax + b)(c + dx) = (a + bx).(cx + d)
⇒ \(a c x+b c+a d x^2+b d x=a c x+b c x^2+a d+b d x\)
⇒ \(a d x^2-b c x^2+b d x-b d x+a c x-a c x=a d-b c\)
⇒ \(x^2 \cdot(a d-b c)=(a d-b c)\)
∴ \(x^2=\frac{a d-b c}{a d-b c}=1\)
∴ x = ±1
∴ x = +1
9. \((2 x+1)+\frac{3}{2 x+1}=4\), x≠ -1/2
Solution: \((2 x+1)+\frac{3}{2 x+1}=4\)
⇒ \(\frac{(2 x+1)^2+3}{2 x+1}=4\)
⇒ \(4 x^2+4 x+1+3=8 x+4\)
⇒ \(4 x^2+4 x-8 x+4-4=0\)
⇒ \(4 x^2-4 x=0\)
⇒ 4x(x – 1) = 0
∴ Either 4x = 0
∴ x = 0
Or, x – 1 = 0
∴ x = 1
10. \(\frac{x+1}{2}+\frac{2}{x+1}=\frac{x+1}{3}+\frac{3}{x+1}-\frac{5}{6}\), x≠ -1
Solution: \(\frac{x+1}{2}+\frac{2}{x+1}=\frac{x+1}{3}+\frac{3}{x+1}-\frac{5}{6}\)
⇒ \(\frac{2}{x+1}-\frac{3}{x+1}=\frac{x+1}{3}-\frac{x+1}{2}-\frac{5}{6}\)
⇒ \(\frac{2-3}{x+1}=\frac{2(x+1)-3(x+1)-5}{6}\)
⇒ \(\frac{-1}{x+1}=\frac{2 x+2-3 x-3-5}{6}\)
⇒ \(-\frac{1}{x+1}=\frac{-x-6}{6}\)
∴ \(-\frac{1}{x+1}=\frac{-(x+6)}{6}\)
\(x^2+7 x+6=6\)⇒ \(x^2+7 x=0\)
⇒ x(x + 7) = 0
∴ Either x = 0
or, x + 7 = 0
∴ x = -7
11. \(\frac{x+3}{x-3}+\frac{6(x-3)}{x+3}=5\), x≠ 3 , -3
Solution: \(\frac{x+3}{x-3}+\frac{6(x-3)}{x+3}=5\)
or, \(\frac{(x+3)^2+6(x-3)^2}{(x-3)(x+3)}=5\)
or, \(x^2+6 x+9+6\left(x^2-6 x+9\right)=5\left(x^2-9\right)\)
or, \(7 x^2-30 x+9+54-5 x^2+45=0\)
or, \(2 x^2-30 x+108=0\)
or, \(x^2-15 x+54=0\)
or, \(x^2-9 x-6 x+54=0\)
or, x(x – 9) – 6(x – 9) = 0
or, (x – 9)(x – 6) = 0
Either x – 9 =0
∴ x = 9
Or, x – 6 = 0
∴ x = 6
12. \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\), x≠ 0 , -(a+b)
Solution: \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)
or, \(\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}\)
or, \(\frac{x-(a+b+x)}{x(a+b+x)}=\frac{b+a}{a b}\)
or, \(\frac{x-a-b-x}{x^2+a x+b x}=\frac{a+b}{a b}\)
or, \(\frac{-(a+b)}{x^2+a x+b x}=\frac{a+b}{a b}\)
or, \(x^2+a x+b x=-a b\)
or, \(x^2+a x+b x+a b=0\)
or, x(x + a) + b(x + a) = 0
or, (x + a)(x + b) = 0
Either x + a = 0
∴ x = -a
or, x + b = 0
∴ x = -b.
13. \(\frac{1}{x}-\frac{1}{x+b}=\frac{1}{a}-\frac{1}{a+b}\), x ≠ 0 , -b
Solution: \(\frac{1}{x}-\frac{1}{x+b}=\frac{1}{a}-\frac{1}{a+b}\)
or, \(\frac{1}{x}-\frac{1}{a}=\frac{1}{x+b}-\frac{1}{a+b}\)
or, \(\frac{a-x}{a x}=\frac{(a+b)-(x+b)}{(x+b)(a+b)}\)
or, \(\frac{a-x}{a x}=\frac{a+b-x-b}{a x+a b+b x+b^2}\)
or, \(\frac{a-x}{a x}=\frac{a-x}{a x+b x+a b+b^2}\)
or, \(\frac{a-x}{a x}-\frac{a-x}{x(a+b)+b(a+b)}=0\)
\((a-x)\left[\frac{1}{a x}-\frac{1}{x(a+b)+b(a+b)}\right]=0\)Either a – x = 0
∴ x = a
or, \(\frac{1}{a x}-\frac{1}{(a+b) x+b(a+b)}=0\)
or, (a + b)x + (a + b)b = ax
or, \(a x+b x+a b+b^2-a x=0\)
∴ \(b x=-a b-b^2\)
∴ bx = -b(a + b)
∴ x = -(a + b) & x = a
14. \(\frac{1}{(x-1)(x-2)}-\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\), x≠ 1,2,3,4
Solution: \(\frac{1}{(x-1)(x-2)}-\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\)
or, \(\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{1}{6}\)
or, \(\frac{1}{x-4}-\frac{1}{x-1}=\frac{1}{6}\)
or, \(\frac{(x-1)-(x-4)}{(x-4)(x-1)}=\frac{1}{6}\)
or, \(\frac{x-1-x+4}{x^2-5 x+4}=\frac{1}{6}\)
∴ \(x^2 – 5x + 4 = 18\)
or, \(x^2-5 x+4-18=0\)
or, \(x^2-5 x-14=0\)
or, \(x^2-7 x+2 x-14=0\)
or, (x – 7) + 2(x – 7) = 0
or, (x – 7)(x + 2) = 0
Either x – 7 = 0
x = 7
or, x + 2 = 0
x = -2
15. a/x-a + b/x-b = 2c/x-c , x≠ a b,c
Solution: \(\frac{a}{x-a}-\frac{c}{x-c}=\frac{c}{x-c}-\frac{b}{x-b}\)
or, \(\frac{a(x-c)-c(x-a)}{(x-a)(x-c)}=\frac{c(x-b)-b(x-c)}{(x-c)(x-b)}\)
or, \(\frac{a x-a c-c x-a c}{x-a}=\frac{c x-b c-b x+b c}{x-b}=0\)
or, \(\frac{x(a-c)}{x-a}-\frac{x(c-b)}{x-b}=0\)
or, \(x\left[\frac{a-c)}{x-a}-\frac{(c-b)}{x-b}\right]=0\)
∴ Either x = 0 (1)
or, \(\frac{a-c}{x-a}=\frac{c-b}{x-b}\)
(a – c)(x – b) = (c – b)(x – a)
ax – cx – ab + bc = cx – bx – ac + ab
ax – cx – cx + bx = ab + ab – ac – bc
x(a + b – 2c) = 2ab – av – bc
∴ \(x=\frac{2 a b-a c-b c}{a+b-2 c} \& x=0\)
16. x2-(√3+2)x+2√3-0
Solution: \(x^2-\sqrt{3} x-2 x+2 \sqrt{3}=0\)
or, x(X – √3) – 2(x – √3) = 0
or, (x – √3)(x – 2) = 0
∴ Either x – √3 = 0
∴ x = √3
Or, x – 2 = 0
∴ x = 2.
Question 17. The unit digit of a two-digit number exceeds the tens digit by 6 and the product of two digits is less by 12 than the number, let us write by calculating the possible unit digit of the two-digit number.
Solution: Let in a two digit number the digit in the tenth place is x.
∴ Digit in the unit place is x + 6.
∴ The two digited number is 10x + x + 6.
According to the given problem,
x(x + 6) + 12 = 10x + x + 6
or, \(x^2+6 x+12=11 x+6\)
or, \(x^2+6 x-11 x+12-6=0\)
or, \(x^2-5 x+6=0\)
or, \(x^2-3 x-2 x+6=0\)
or, x(x – 3) – 2(x – 3) = 0
∴ (x – 3)(x – 2) = 0