WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.1
Question 1: write the quadratic polynomials from the following polynomials by understating them.
- x2 – 7x+2
Solution: It is a quadratic equation. - 7x3– x(x+2)
= 7x3– x2 – 2x
Solution: It is not a quadratic equation. - 2x(x+5)+1
=2x2+10x + 1
Solution: It is a quadratic equation. - 2x – 1
Solution: It is not a quadratic equation.
Question 2: Which of the following equations can be written in the form of ax2+ bx + c = 0, where a, b, c are real numbers and a 0?
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⇒ \(x-1+\frac{1}{x}=6,(x \neq 0)\)
Solution: \(\frac{x^2-x+1}{x}=6\)
⇒ \(x^2-x+1=6 x\)
⇒ \(x^2-x-6 x+1=0\)
⇒ \(x^2-7 x+1=0\)
This is a quadratic equation in the general form: ax2 + bx + c = 0
1. x +3/x = x2, (x ≠ 0)
Solution: \(\frac{x^2+3}{x}=x^2\)
⇒ \(x^2+3=x^3\)
⇒ \(x^3-x^2-3=0\)
This is not a quadratic equation.
2. x2-6√x+2=0
Solution: This is not a quadratic equation.
3. (x-2)2= x2-4x+4
Solution: This is not a quadratic equation; it is an identity.
Question 3. Let us determine the power of the variable for which the equation x – x3– 2 = 0 will become a quadratic equation.
Solution: x6-x3-2=0
=>(x3)2-1.x3-2=0
This is a quadratic equation with respect to x3.
Question 4. 1. Let us determine the value of ‘a’ for which the equation (a-2)x2+ 3x + 5 = 0 will not be a quadratic equation.
Solution:
Given
(a-2)x2+3x+5=0.
If the value of a = 2, it will be 0x2 + 3x + 5 = 0
=> 3x + 5 = 0
So, if the value of a = 2, it will not be a quadratic equation. X
2. If x/4-x=1/3x(x =not 0, x =not 4) be expressed in the form of ax2 + bx + c = (a =not 0), then4-x 3x let us determine the co-efficient of x.
Solution: \(\frac{x}{4-x}=\frac{1}{3 x}\)
⇒ \(\frac{x}{4-x}-\frac{1}{3 x}=0\)
⇒ \(\frac{3 x^2-(4-x)}{(4-x) 3 x}=0\)
⇒ \(3 x^2-4+x=0\)
⇒ \(3 x^2+x-4=0\)
3. Let us express 3x2 + 7x + 23 = (x + 4) (x+3)+ 2 in the form of the quadratic equation ax2 + bx + c = 0 (a ≠ 0)
Solution: \(3 x^2+7 x+23=(x+4)(x+3)+2\)
⇒ \(3 x^2+7 x+23=x^2+7 x+12+2\)
⇒ \(3 x^2+7 x-7 x+23-14-x^2=0\)
⇒ \(2 x^2+0 \cdot x+9=0\)
4. Let us express the equation (x+2)3= x(x2-1) in the form of ax2 + bx + c = 0 (a = not0) and write the coefficients of x2, x, and x°.
Solution: \((x+2)^3=x\left(x^2-1\right)\)
⇒ \(x^3+6 x^2+12 x+8=x^3-x\)
⇒ \(x^3-x^3+6 x^2+12 x+x+8=0\)
⇒ \(6 x^2+13 x+8=0\)
∴ Co-efficient of x2 is 6,
Co-efficient of x is 13,
The coefficient of x0 is 8.
Question 5. Let us construct quadratic equations in one variable from the following statements.
1. Divide 42 into two parts such that one part is equal to the square of the other part.
Solution: Let one part be x and the other part is 42 – x.
∴ \(x^2=42-x\)
⇒ \(x^2+x-42=0\)
Where x is the first part.
2. The product of two consecutive positive odd numbers is 143.
Solution: Let one positive odd number be 2x – 1 & the next odd number is 2x + 1.
∴ (2x – 1)(2x + 1) = 143
⇒ \(4 x^2-1=143\)
⇒ \(4 x^2-1-143=0\)
⇒ \(4 x^2-144=0\)
⇒ \(x^2-36=0\)
3. The sum of the squares of two consecutive numbers is 313.
Solution: Let one number be x and the other number be x + 1.
According to the given conditions,
⇒ \((x)^2+(x+1)^2=313\)
⇒ \(x^2+x^2+2 x+1-313=0\)
⇒ \(2 x^2+2 x-312=0\)
⇒ \(2\left(x^2+x-156\right)=0\)
∴ \(x^2+x-156=0\)
Question 6. Let us construct the quadratic equations in one variable from the following statements.
1. The length of the diagonal of a rectangular area is 15 m and the length exceeds its breadth by 3 m
Solution:
Given
The length of the diagonal of a rectangular area is 15 m and the length exceeds its breadth by 3 m
Let the breath and length of the rectangle be x m. and (x+3) m respectively.
According to the given problem,
⇒ \(x^2+(x+3)^2=(15)^2\)
⇒ \(x^2+x^2+6 x+9=225\)
⇒ \(2 x^2+6 x+9-225=0\)
⇒ \(2 x^2+6 x-216=0\)
⇒ \(2\left(x^2+3 x-108\right)=0\)
∴ \(x^2+3 x-108=0\)
2. One person bought some kg sugar for Rs. 80. If he could get 4 kg more sugar with that money, then the price of 1 kg sugar would be less by Rs. 1.
Solution:
Given
One person bought some kg sugar in Rs. 80. If he would get 4 kg more sugar with that money,
Let the price of x kg. sugar is Rs.80
∴ The price of 1 kg sugar is Rs. \(\frac{80}{x}\)
∴ According to the given problem,
⇒ \(\left(\frac{80}{x}-1\right) \times(x+4)=80\)
⇒ \(80-x+\frac{320}{x}-4-80=0\)
⇒ \(-x^2+320-4 x=0\)
∴ \(x^2+4 x-320=0\) is the required equation.
3. The distance between the two stations is 300 km. A train went to the second station from the first station with uniform velocity. If the velocity of the train is 5 km/hour more, then the time taken by the train to reach the second station would be less by 2 hours.
Solution: Let the speed of the train be x m/hr.
∴ To go 300 km the train takes \(\frac{300}{x} \mathrm{hr}.\)
If the speed of the train is (x + 5) km/hr, the time required will be \(\frac{300}{x+5} \mathrm{hr}\)
According to the problem,
⇒ \(\frac{300}{x}-\frac{300}{x+5}=2\)
⇒ \(\frac{300(x+5)-300(x)}{x(x+5)}=2\)
⇒ \(300 x+1500-300 x=2\left(x^2+5 x\right)\)
⇒ \(1500=2\left(x^2+5 x\right)\)
⇒ \(x^2+5 x-750=0\)
4. A clock seller sold a clock by purchasing it at Rs. 336. The amount of his profit percentage is as much as the amount with which he bought the clock.
Solution: Let the cost price of the watch is Rs. x.
∴ Profit = x% of x = \(\text { Rs. } x \times \frac{x}{100}=\text { Rs. } \frac{x^2}{100}\)
According to the problem,
⇒ \(x+\frac{x^2}{100}=336\)
⇒ \(100 \mathrm{x}+\mathrm{x}^2=33600\)
⇒ \(x^2+100 x-33600=0\)
5. If the velocity of the stream is 2 km/hr, then the time taken by Ratanmajhi to cover 21 km downstream and upstream is 10 hours.
Solution: Let the speed of the boat in still water is x km/hr.
∴ Speed of the boat with the current is (x + 2)km/hr and speed of the boat against the current is (x – 2)km/hr.
According to the problem,
⇒ \(\frac{21}{x+2}+\frac{21}{x-2}=10\)
⇒ \(\frac{21(x-2)+21(x+2)}{(x+2)(x-2)}=10\)
⇒ 21x – 42 + 21x + 42 = 10(x + 2)(x – 2)
⇒ \(42 x=10\left(x^2-4\right)\)
⇒ \(21 x=5\left(x^2-4\right)\)
⇒ \(-5 x^2+21 x+20=0\)
∴ \(5 x^2-21 x-20=0\) is the required equation.
6. The time taken to clean out the garden of Majid is 3 hours more than Mahim. Both of them together can complete the work in 2 hours.
Solution: Let Mahim alone take x hrs. to finish the work & Majid alone takes (x + 3) hrs to finish the work.
According to the problem,
⇒ \(\left(\frac{1}{x}+\frac{1}{x+3}\right) \times 2=1\)
⇒ \(\left(\frac{x+3+x}{x(x+3)}\right) \times 2=1\)
⇒ \(4 x+6=x^2+3 x\)
⇒ \(x^2+3 x-4 x-6=0\)
∴ \(x^2-x-6=0\) is the required equation.
Question 7. The unit digit of a two-digit number exceeds its ten’s digit by 6 and the product of two digits is less by 12 from the number.
Solution: Let in a two-digit number, the digit in the tenth place is x and the digit 2 unit place is (x + 6)
∴ The number is 10x + (x + 6)
According to the given problem,
x(x + 6) + 12 = 10x + x + 6
⇒ \(x^2+6 x+12=11 x+6\)
⇒ \(x^2-5 x+6=0\) is the required equation.
Question 8. There is a road of equal width around the outside of a rectangular playground having a length of 45 m and breadth of 40 m and the area of the road is 450 sq.m.
Solution: Let the width of the path = xm.
According to the problem,
⇒ (45 + 2x) x (40 + 2x) – 45 x 40 = 450
⇒ \(1800 + 90x + 80x + 4 x^2 – 1800 = 450\)
⇒ \(4 x^2+170 x-450=0\)
⇒ \(2 x^2+85 x-225=0\) is the required equation.
Question 9. Let me write by calculating, for what value of k, 1 will be a root of the quadratic equation x2 + kx+3=0.
Solution: As one root of the equation \(x^2+k x+3=0\) is 1
∴ \(1^2+\mathrm{k} \cdot 1+3=0\)
⇒ 1 + k + 3 = 0
∴ k + 4 = 0
∴ k = -4
Question 10. I solve and write the two roots of the quadratic equation \(\frac{a}{a x-1}+\frac{b}{b x-1}=a+b,\left[x \neq \frac{1}{a}, \frac{1}{b}\right]\)
Solution: \(\frac{a}{a x-1}-a+\frac{b}{b x-1}-b=0\)
or, \(\frac{a-a^2 x+a}{a x-1}+\frac{b-b^2 x+b}{b x-1}=0\)
or, \(\left(2 a-a^2 x\right)(b x-1)+\left(2 b-b^2 x\right) \cdot(a x-1)=0\)
or, \(2 a b x-a^2 b x^2-2 a+a^2 x+2 a b x-a b^2 x^2-2 b+b^2 x=0\)
or, \(-a^2 b x^2-a b^2 x^2+2 a b x+a^2 x+2 a b x+b^2 x-(2 a+2 b)=0\)
or, \(-\left(a^2 b+a b^2\right) x+\left(a^2+b^2+4 a b\right) x-(2 a+2 b)=0\)
or, \(\left(a^2 b+a b^2\right) x^2-\left\{(a+b)^2+2 a b\right\} x+(2 a+2 b)=0\)
or, \(a b(a+b) x^2-\left\{(a+b)^2+2 a b\right\} x+2(a+b)=0\)
or, \(a b(a+b) x^2-(a+b)^2 x-2 a b x+2(a+b)=0\)
or, (a + b)x{abx – (a + b)} – 2{abx – (a + b)}
or, {(a + b)x – 2}{abx – (a + b)} = 0
Either (a + b)x – 2 = 0
∴ \(x=\frac{2}{a+b}\)
or, abx – (a + b) = 0
∴ \(x=\frac{a+b}{a b}\)
Question 11. I solve the quadratic equation \(\frac{x+3}{x-3}+\frac{x-3}{x+3}=2 \frac{1}{2}\) (x ≠ -3, 3 ) .
Solution: \(\frac{x+3}{x-3}+\frac{x-3}{x+3}=2 \frac{1}{2}\)
⇒ \(\frac{(x+3)^2+(x-3)^2}{(x+3)(x-3)}=\frac{5}{2}\)
⇒ \(\left(x^2+6 x+9+x^2-6 x+9\right) \times 2=5\left(x^2-9\right)\)
⇒ \(x^2+36=5 x^2-45\)
⇒ \(5 x^2-4 x^2-45-36=0\)
⇒ \(x^2-81=0\)
⇒ \((x)^2-(9)^2=0\)
⇒ (x + 9)(x – 9) = 0
∴ Either x + 9 = 0
∴ x = -9
or, x – 9 = 0
∴ x = 9.