(1) The digit in the unit place of a two digit number is 5 more than the digit in the tens place. The product of the digits is 14 less than the number. Express the statement in quadratic equation. Find also the number.
(2) Express the equation (x+2)³= x(x²-1) in the quadratic equation of the form ax²+bx+c=0. Then write the coefficient of x²,x and x⁰.
(3) The sum of the squares of two consecutive integers is 265. Express the statement in an equation.
(4) The diagonals of a rectangular field is 25 metre. The length is 5 metre more than its breadth-Express the statement in a quadratic equation and find the breadth of the rectangle.
Answers
Answer:
Question :-
1) The digit in the unit place of a two digit number is 5 more than the digit in the tens place. The product of the digits is 14 less than the number. Express the statement in quadratic equation. Find also the number.
Solution :-
Let the digits in the unit place be x and in the tens place be y.
Then the number = 10y + x.
According to the question,
➪ x = y + 5
➻ (10y + x) - xy = 14
➻ 10y + y + 5 - (y + 5)y = 14
➻ 10y + y + 5 - y² - 5y - 14 = 0
➻ 10y + y - 5y - y² + 5 - 14 = 0
➻ 6y - y² - 9 = 0
➻ - y² + 6y - 9 = 0
➻ - (y² - 6y + 9) = 0
➻ y² - 6y + 9 = 0, a quadratic equation.
➻ y² - (3 + 3)y + 9 = 0
➻ y² - 3y - 3y + 9 = 0
➻ y(y - 3) - 3(y - 3) = 0
➻ (y - 3) (y - 3) = 0
➻ (y - 3)² = 0
➻ y = 3
➻ x = y + 5
➻ x = 3 + 5
∴ x = 8
➻ The number of = 10y + x
➻ The number of = 10 × 3 + 8
➻ The number of = 30 + 8
∴ The number of = 38
Hence, the number is 38. (Ans)
Question :-
2) Express the equation (x + 2)³ = x(x² - 1) in the quadratic equation of the form ax² + bx + c = 0. Then write the coefficient of x², x and x⁰.
Solution :-
➪ (x + 2)³ = x(x² - 1)
➻ (x)³ + 3 × (x)² × 2 + 3 × x × (2)² + (2)³ = x³ - x
➻ x³ + 6x² + 12x + 8 = x³ - x
➻ x³ - x³ + 6x² + 12x + x + 8 = 0
➻ 6x² + 13x + 8 = 0 [ Which is a quadratic equation of the form ax² + bx + c = 0 ]
Here, coefficient of x² is 6, coefficient of x is 13 and the coefficient of x⁰ is 8 (Ans).
Question :-
3) The sum of the squares of two consecutive integers is 265. Express the statement in an equation.
Solution :-
Let the two consecutive integers be x and x + 1
According to the question,
➪ (x)² + (x + 1)² = 265
➻ x² + (x)² + 2 × x × 1 + (1)² = 265
➻ x² + x² + 2x + 1 = 265
➻ 2x² + 2x + 1 = 265
➻ 2x² + 2x + 1 - 265 = 0
➻ 2x² + 2x - 264 = 0
➻ 2(x² + x - 132) = 0
➻ x² + x - 132 = 0 [ Which is the required equation ]
∴ The required equation is x² + x - 132
Question :-
4) The diagonals of a rectangular field is 25 metre. The length is 5 metre more than its breadth - Express the statement in a quadratic equation and find the breadth of the rectangle.
Solution :-
Let the breadth of the rectangle be x metre.
Then the length, of the rectangle be ( x + 5) metre.
Now, since length² + breadth² = diagonal²
∴ (x + 5)² + x² = 25² [ diagonal = 25 m given ]
➻ (x)² + 2 × x × 5 + (5)² + x² = 25 × 25
➻ x² + 10x + 25 + x² = 625
➻ x² + x² + 10x + 25 - 625 = 0
➻ 2x² + 10x - 600 = 0
➻ 2(x² + 5x - 300) = 0
➻ x² + 5x - 300 = 0 [ which is the required quadratic equations ]
Now, to solve the equation,
➻ x² + (20 - 15)x - 300 = 0
➻ x² + 20x - 15x - 300 = 0
➻ x(x + 20) - 15(x + 20) = 0
➻ (x + 20) (x - 15) = 0
➻ (x + 20) = 0
➻ x + 20 = 0
➻ x = - 20
Either,
➻ ( x - 15) = 0
➻ x - 15 = 0
➻ x = 15
∴ The breadth of the rectangle = 15 metre.
Formula used :-
➔ (a + b)²
➭ a² + 2ab + b²
➔ (a + b)³
➭ a³ + 3a²b + 3ab² + b³
answer
the number 10y+x. By the problem, x = y +5
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14or, 10y+y+5-(+5)=14 or, 10y+y+5-2-Sy-14-0
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14or, 10y+y+5-(+5)=14 or, 10y+y+5-2-Sy-14-0or,-2+6y-9-0
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14or, 10y+y+5-(+5)=14 or, 10y+y+5-2-Sy-14-0or,-2+6y-9-0or, 2-6y+9=0, a quadratic equation.
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14or, 10y+y+5-(+5)=14 or, 10y+y+5-2-Sy-14-0or,-2+6y-9-0or, 2-6y+9=0, a quadratic equation.or, (y-3)2-0 or, y=3..
the number 10y+x. By the problem, x = y +5and (10y+x)-xy = 14or, 10y+y+5-(+5)=14 or, 10y+y+5-2-Sy-14-0or,-2+6y-9-0or, 2-6y+9=0, a quadratic equation.or, (y-3)2-0 or, y=3..x=8.