Question

(1) The product of two consecutive positive odd integers is 99. What are the number ????
(2) The difference of two positive integral is 2 and the sum of their squares is 74. What are the numbers ????
(3) From the quadratic equation whose roots are -11 and -9.
(4) A number is less than its square by 20. Express the statement in a quadratic equation. Find also the discriminant from the equation.
(5) The sum of the roots of a quadratic equation is (2p + 1) and their product is 7p. The equation is _______.
(6) The roots of a quadratic equation are 6 and -7. So the equation is _____.
(7) The sum of the roots of the equation x2-4x=k(x-1)-5 is 7, the value of k is_____.
(8) 2 is a root of both the equations x2+bx+12=0andx2+bx+c=0,then find the value of c.
(9) A man travels first 30 km towards east, then 25km towards north and 6km towards west and 7km towards south to reach the destination. How far is the destination from the starting point.​

Answer 1

Answer:

Question :-

1) The product of two consecutive positive odd integers is 99. What are the number?

Solution :-

Let the number be x and (x + 2)

∴ By the problem

➪ x(x + 2) = 99

➭ x² + 2x - 99 = 0

➭ x² + (11 - 9)x - 99 = 0

➭ x² + 11x - 9x - 99 = 0

➭ x(x + 11) - 9(x + 11) = 0

➭ (x + 11) (x - 9) = 0

➭ (x + 11) = 0

➭ x + 11 = 0

➭ x = - 11

Either,

➭ ( x - 9) = 0

➭ x - 9 = 0

➭ x = 9

∴ x = 9, [ $\becsuse$ x is only positive ]

∴ The numbers are 9, and - 11.

Question :-

2) The difference of two positive integral numbers is 2 and the sum of their squares is 74. What are the numbers?

Solution :-

Let the smaller number be x.

∴ The greater number is x + 2

∴ By the problem

➪ (x + 2)² + x² = 74

➭ (x)² + 2 × x × 2 + (2)² + x² = 74

➭ x² + 4x + 4 + x² = 74

➭ x² + x² + 4x + 4 - 74 = 0

➭ 2x² + 4x - 70 = 0

➭ 2(x² + 2x - 35) = 0

➭ x² + 2x - 35 = 0

➭ x² + (7 - 5)x - 35 = 0

➭ x² + 7x - 5x - 35 = 0

➭ x(x + 7) - 5(x + 7) = 0

➭ (x + 7) (x - 5) = 0

➭ (x + 7) = 0

➭ x + 7 = 0

➭ x = - 7

Either,

➭ (x - 5) = 0

➭ x - 5 = 0

➭ x = 5

Neglecting the negative value,

The smaller number = 5

and the greater number = 5 + 2 = 7.

∴ The numbers are 7.

Question :-

3) From the quadratic equation whose roots are - 11 and - 9.

Solution :-

Here, sum of the roots

➪ - 11 - 9

➭ - 20

Product of the roots

➪ (- 11) × (- 9)

➭ 99

∴ The required quadratic equation:

x² - (sum of the roots)x + Product of the roots = 0

∴ x² - (- 20)x + 99 = 0

➭ x² + 20x + 99 = 0

∴ The quadratic equation x² + 20x + 99 = 0

Question :-

4) A number is less than its square by 20. Express the statement in a quadratic equation. Find also the discriminant from the equation.

Solution :-

Let the number be x

∴ By the problem

➪ x² - x = 20

➭ x² - x - 20 = 0 is the required quadratic equation. [ Comparing with the general from ax² + bx + c = 0, here a = 1, b = - 1, c = - 20. ]

Now, the discriminant

➪ (- 1)² - 4(1)(- 20)

➭ 1 + 80

➭ 81

∴ The discriminant from the equation is 81

Question :-

5) The sum of the roots of a quadratic equation is (2p + 1) and their product is 7p. The equation is ______.

Solution :-

∴ The quadratic equation is

x² - (sum of the roots)x + Product of the roots = 0

∴ x² + (2p + 1)x + 7p = 0

∴ The equation is x² + (2p + 1)x + 7p = 0

Question :-

6) The roots of a quadratic equation are 6 and - 7. So the equation is _____.

Solution :-

The quadratic equation is

x² - (sum of the roots)x + Product of the roots = 0

➪ x² - (6 - 7)x + 6 × (- 7) = 0

➭ x² + x - 42 = 0

∴ The equation is x² + x - 42 = 0

7) The sum of the roots of the equation x² - 4x = k(x - 1) - 5 is 7, then the value of k is ______.

Solution :-

$\because$ x² - 4x = k(x - 1) - 5

➪ x² - 4x = kx - k - 5

➪ x² - 4x - kx + k + 5 = 0

➪ x² - x(4 + k) + (k + 5) = 0

Now by the problem,

➪ 4 + k = 7

➪ k = 7 - 4

➪ k = 3

∴ The value of k is 3

Question :-

8) 2 is a roots of both the equations x² + bx + 12 = 0 and x² + bx + c = 0, then find the value of c.

Solution :-

Since, 2 is a root of x² + bx + 12 = 0

∴ 2² - b × 2 + 12 = 0

➭ 4 + 2b + 12 = 0

➭ 2b + 12 + 4 = 0

➭ 2b + 16 = 0

➭ 2b = - 16

➭ b = - 16/2

➭ b = - 8

Now, 2 is a root of x² + bx + c = 0

∴ 2² + (- 8) × 2 + c = 0

➭ 4 - 16 + c = 0

➭ c = 16 - 4

➭ c = 12

∴ The value of c is 12.

9) A man travels first 30 km towards east, then 25 km towards north and 6 km towards west and 7 km towards south to reach the destination. How far is the destination from the starting point.

Solution :-

According to the diagram,

O is the starting point,

OB = 30 km towards east

BC = 25 km towards north

CD = 6 km towards west

DP = 7 km towards south.

∴ P is the destination, the end point.

∴ In ∆OAP, by the Pythagoras theorem,

OP² = OA² + AP²

= (OB - BA)² + (AD - DP)²

= (OB - CD)² + (BC - DP)²

= (30 - 6)² + (25 - 7)²

= 24² + 18²

= 576 + 324

= 900

= 30²

∴ OP = 30 KM

∴ That is the destination is at a distance 30 km from the starting point.