1. Write down which / which of the following polynomial expressions are quadratic polynomial expressions.
(i) x² + 7x + 2 = 0
Answer:
The highest power of x in the polynomial expression is 2
Therefore, this is a quadratic polynomial expression.
(ii) 7x⁵ - x(x + 2)
Answer:
The highest power of x in the polynomial expression is 5
Therefore, this is not a quadratic polynomial expression.
(iii) 2x(x + 5) + 1
Answer:
2x(x + 5) + 1 = 2x² + 10x + 1
The highest power of x in the polynomial expression is 2
Therefore, this is a quadratic polynomial expression.
(iv) 2x - 1
Answer:
The highest power of x in the polynomial expression is 1
Therefore, this is not a quadratic polynomial expression.
2. Write which of the following equations can be written in the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
(i) \(x - 1 + {1 \over x} = 0, (x \neq 0)\)
Answer:
\(x - 1 + {1 \over x} = 0\)
Or, \({{x² - x + 1} \over x} = 0\)
Or, x² - x + 1 = 0
This equation can be written in the form ax² + bx + c = 0 (where a, b, c are real numbers and a ≠ 0).
(ii) \(x + {3 \over x} = x², (x \neq 0)\)
Answer:
\(x + {3 \over x} = x², (x \neq 0)\)
Or, 3x² = 4 - x
Or, 3x² + x - 4 = 0
This equation can be written in the form ax² + bx + c = 0 (where a, b, c are real numbers and a ≠ 0).
(iii) x² - 6√x + 2 = 0
Answer:
This equation cannot be written in the form ax² + bx + c = 0 (where a, b, c are real numbers and a ≠ 0).
(iv) (x - 2)² = x² - 4x + 4
Answer:
This is an identity.
Therefore, this equation cannot be written in the form ax² + bx + c = 0 (where a, b, c are real numbers and a ≠ 0).
3. Determine with respect to which power of the variable is the equation x⁶ - x³ - 2 = 0 a quadratic equation.
Answer:
x⁶ - x³ - 2 = 0
Or, (x³)² - x³ - 2 = 0
This equation is a quadratic equation with respect to x³
That is, the equation is a quadratic equation with respect to the power 3 of the variable x.
4.(i) For what value of a will the equation (a - 2)x² + 3x + 5 = 0 not be a quadratic equation, determine it.
Answer:
The equation (a - 2)x² + 3x + 5 = 0 will not be a quadratic equation if the coefficient of x² is 0.
That is, if a - 2 = 0 or a = 2, the equation (a - 2)x² + 3x + 5 = 0 will not be a quadratic equation.
(ii) Express \({x \over {4 - x}} = {1 \over 3x}\) in the form of the quadratic equation ax² + bx + c = 0, (a ≠ 0) and determine what the coefficient of x will be.
Answer:
\({x \over {4 - x}} = {1 \over 3x}\)
Or, 3x² = 4 - x
Or, 3x² + x - 4 = 0
∴ When the given equation is expressed in the form of a quadratic equation, the coefficient of x will be 1
(iii) Express 3x² + 7x + 23 = (x + 4)(x + 3) + 2 in the form of the quadratic equation ax² + bx + c = 0, (a ≠ 0).
Answer:
3x² + 7x + 23 = (x + 4)(x + 3) + 2
Or, 3x² + 7x + 23 = x² + 4x + 3x + 12 + 2
Or, 3x² + 7x + 23 = x² + 7x + 14
Or, 3x² + 7x + 23 - x² - 7x - 14 = 0
Or, 2x² + 0·x + 9 = 0
∴ Expressing the given equation in the form of a quadratic equation gives 2x² + 0·x + 9 = 0
(iv) Express the equation (x + 2)³ = x(x² - 1) in the form of the quadratic equation ax² + bx + c = 0, (a ≠ 0) and write the coefficients of x², x, and x⁰.
Answer:
(x + 2)³ = x(x² - 1)
Or, x³ + 3·x²·2 + 3·x·2² + 2³ = x³ - x
Or, x³ + 6x² + 12x + 8 - x³ + x = 0
Or, 6x² + 13x + 8 = 0
∴ Expressing the given equation in the form of a quadratic equation gives 6x² + 13x + 8 = 0
Coefficient of x² is 6
Coefficient of x is 13
Coefficient of x⁰ is 8
5. Formulate univariate quadratic equations from the following statements.
(i) Divide 42 into two parts such that one part is equal to the square of the other part.
Answer:
Let one part be x
∴ The other part is x²
According to the condition, x² + x = 42
Or, x² + x - 42 = 0
∴ The required univariate quadratic equation is x² + x - 42 = 0
(ii) The product of two consecutive positive odd numbers is 143
Answer:
Let the two consecutive positive odd numbers be (2x - 1) and (2x + 1)
According to the condition, (2x - 1)(2x + 1) = 143
Or, (2x)² - 1² = 143
Or, 4x² - 1 - 143 = 0
Or, 4x² - 144 = 0
Or, 4(x² - 36) = 0
Or, x² - 36 = 0
∴ The required univariate quadratic equation is x² - 36 = 0
(iii) The sum of the squares of two consecutive numbers is 313
Answer:
Let the two consecutive numbers be x and x + 1
According to the condition, x² + (x + 1)² = 313
Or, x² + x² + 2x + 1 - 313 = 0
Or, 2x² + 2x - 312 = 0
Or, 2(x² + x - 156) = 0
Or, x² + x - 156 = 0
∴ The required univariate quadratic equation is x² + x - 156 = 0
6. Formulate univariate quadratic equations from the following statements.
(i) The length of the diagonal of a rectangular field is 15 meters and its length is 3 meters more than its width.
Answer:
Let the width of the rectangular field = x meters
∴ The length of the rectangular field = (x + 3) meters
∴ The length of the diagonal of the rectangular field = \(\sqrt{x^2 + (x + 3)^2}\) meters
According to the condition, \(\sqrt{x^2 + (x + 3)^2}\) = 15
Or, x² + x² + 6x + 9 = 15²
Or, 2x² + 6x + 9 = 225
Or, 2x² + 6x + 9 - 225 = 0
Or, 2x² + 6x - 216 = 0
Or, 2(x² + 3x - 108) = 0
Or, x² + 3x - 108 = 0
∴ The required univariate quadratic equation is x² + 3x - 108 = 0
(ii) A person bought some kg of sugar for 80 rupees. If he had gotten 4 kg more sugar for that amount, the price of sugar per kg would have been 1 rupee less.
Answer:
Let him have bought x kg of sugar for 80 rupees.
∴ Price of 1 kg of sugar = \({80 \over x}\) rupees
If he had gotten 4 kg more sugar for that amount, the price per kg of sugar would have been = \({80 \over x + 4}\) rupees
According to the condition, \({80 \over x} - {80 \over x + 4} = 1\)
Or, \({80(x + 4) - 80x \over x(x + 4)} = 1\)
Or, \({80x + 320 - 80x \over x^2 + 4x} = 1\)
Or, x² + 4x = 320
Or, x² + 4x - 320 = 0
∴ The required univariate quadratic equation is x² + 4x - 320 = 0
(iii) The distance between two stations is 300 km. A train went from the first station to the second station at a uniform speed. If the speed of the train were 5 km/hr more, it would have taken 2 hours less to reach the second station.
Answer:
Let the speed of the train be x km/hr
Time taken by the train to go 300 km is \({300 \over x}\) hours
If the speed of the train were 5 km/hr more, the time taken to go 300 km would be \({300 \over x + 5}\) hours
According to the condition, \({300 \over x} - {300 \over x + 5} = 2\)
Or, \({300(x + 5) - 300x \over x(x + 5)} = 2\)
Or, \({300x + 1500 - 300x \over x^2 + 5x} = 2\)
Or, 2(x² + 5x) = 1500
Or, x² + 5x = 750
Or, x² + 5x - 750 = 0
∴ The required univariate quadratic equation is x² + 5x - 750 = 0
(iv) A watch seller bought a watch and sold it for 336 rupees. His percentage profit was equal to the purchase price of the watch in rupees.
Answer:
Let him have bought the watch for x rupees.
∴ His profit is x%
∴ Selling price of the watch = \(x + x \times {x \over 100}\) rupees
According to the condition, \(x + {x^2 \over 100} = 336\)
Or, \({100x + x^2 \over 100} = 336\)
Or, 100x + x² = 33600
Or, x² + 100x - 33600 = 0
∴ The required univariate quadratic equation is x² + 100x - 33600 = 0
(v) If the speed of the current is 2 km/hr, it takes Ratan Majhi 10 hours to go 21 km downstream and return back that distance.
Answer:
Let the speed of the boat in still water = x km/hr
∴ Speed of the boat downstream = (x + 2) km/hr
And speed of the boat upstream = (x - 2) km/hr
∴ Time taken to go 21 km downstream is \({21 \over x + 2}\) hours
And time taken to go 21 km upstream is \({21 \over x - 2}\) hours
According to the condition, \({21 \over x + 2} + {21 \over x - 2} = 10\)
Or, \({21(x - 2) + 21(x + 2) \over (x + 2)(x - 2)} = 10\)
Or, \({21x - 42 + 21x + 42 \over x^2 - 4} = 10\)
Or, 10(x² - 4) = 42x
Or, 10x² - 40 - 42x = 0
Or, 2(5x² - 21x - 20) = 0
Or, 5x² - 21x - 20 = 0
∴ The required univariate quadratic equation is 5x² - 21x - 20 = 0
(vi) It takes Majid 3 hours more than Mohim to clean our house garden. Both of them together can finish the work in 2 hours.
Answer:
Let Mohim do the work in x hours
∴ Majid does the work in (x + 3) hours
Mohim does \({1 \over x}\) part of the work in 1 hour
Majid does \({1 \over x + 3}\) part of the work in 1 hour
Together they do \({1 \over x} + {1 \over x + 3}\) part in 1 hour
Together they do \(2({1 \over x} + {1 \over x + 3})\) part in 2 hours
According to the condition, \(2({1 \over x} + {1 \over x + 3}) = 1\)
Or, \({(x + 3) + x \over x(x + 3)} = {1 \over 2}\)
Or, \({2x + 3 \over x^2 + 3x} = {1 \over 2}\)
Or, x² + 3x = 2(2x + 3)
Or, x² + 3x = 4x + 6
Or, x² + 3x - 4x - 6 = 0
Or, x² - x - 6 = 0
∴ The required univariate quadratic equation is x² - x - 6 = 0
(vii) The digit in the unit's place of a two-digit number is 6 more than the digit in the ten's place, and the product of the digits is 12 less than the number itself.
Answer:
Let the ten's digit be x
The unit's digit will be (x + 6)
∴ The number is 10x + (x + 6)
According to the condition, 10x + (x + 6) = x(x + 6) - 12
Or, 11x + 6 = x² + 6x - 12
Or, x² + 6x - 11x - 6 - 12 = 0
Or, x² - 5x - 18 = 0
∴ The required univariate quadratic equation is x² - 5x - 18 = 0
(viii) There is a path of uniform width all around the outside of a rectangular playground that is 45 meters long and 40 meters wide, and the area of that path is 450 square meters.
Answer:
Area of the rectangular playground = 45 × 40 square meters = 1800 square meters
Let the path be x meters wide
Length of the rectangular field including the path = (45 + 2x) meters
Width of the rectangular field including the path = (40 + 2x) meters
∴ Area of the rectangular field including the path = (45 + 2x)(40 + 2x) square meters
∴ Area of the path = {(45 + 2x)(40 + 2x) - 1800} square meters
According to the condition, (45 + 2x)(40 + 2x) - 1800 = 450
Or, 1800 + 90x + 80x + 4x² - 1800 = 450
Or, 4x² + 170x - 450 = 0
Or, 2(2x² + 85x - 225) = 0
Or, 2x² + 85x - 225 = 0
∴ The required univariate quadratic equation is 2x² + 85x - 225 = 0
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