Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 4 Quadratic Equations Important Questions and Answers.

## RBSE Class 10 Maths Chapter 4 Important Questions Quadratic Equations

Objective Type Questions—

Question 1.

The solution of the equation 5x^{2} – \(\frac{49}{5}\) = 0 are—

(A) \(\frac{49}{25}\), \(\frac{-49}{25}\)

(B) \(\frac{5}{7}\), \(\frac{-5}{7}\)

(C) \(\frac{-7}{5}\), \(\frac{7}{5}\)

(D) \(\frac{-7}{\sqrt{5}}\), \(\frac{7}{\sqrt{5}}\)

Answer:

(C) \(\frac{-7}{5}\), \(\frac{7}{5}\)

Question 2.

If one solution of the equation x^{2} + 3ax + k = 0 is x = – a then the value of k will be—

(A) 0

(B) 2a^{2}

(C) a^{2}

(D) -2a

Answer:

(B) 2a^{2}

Question 3.

The square of a number is 70 more than its three times. The equation expressing this statement is—

(A) x^{2} + 3x – 70 = 0

(B) x^{2} – 3x – 70 = 0

(C) x^{2} – 3x + 70 = 0

(D) x^{2} + 3x + 70 = 0

Answer:

(B) x^{2} – 3x – 70 = 0

Question 4.

The roots of the equation (x – 3)^{2} = 3 are—

(A) 3 ± \(\sqrt {3}\)

(B) – 3 ± \(\sqrt {3}\)

(C) 0

(D) 6

Answer:

(A) 3 ± \(\sqrt {3}\)

Question 5.

The roots of the quadratic equation px^{2} + qx + r = 0, p ≠ 0 will be equal if—

(A) p^{2} < 4qr

(B) p^{2} > 4qr

(C) q^{2} = 4pr

(D) p^{2} = 4qr

Answer:

(C) q^{2} = 4pr

Question 6.

The roots of the equation ax^{2} + bx + c = 0, a 0 will not be real if—

(A) b^{2} < 4ac

(B) b^{2} > 4ac

(C) b^{2} = 4ac

(D) b = 4ac

Answer:

(A) b^{2} < 4ac

Question 7.

The solution of the equation x^{2} – 4x = 0 are—

(A) 4, 4

(B) 2, 2

(C) 0, 4

(D) 0. 2

Answer:

(C) 0, 4

Very Short Answer Type Questions—

Question 1.

If 6x^{2} = 54 then write the value of x.

Solution:

x^{2} = \(\frac{54}{6}\) = 9

Hence x = ± 3

Question 2.

Write the solution of the equation x^{2} – \(\frac{x}{3}\) = 0

Solution:

x^{2} – \(\frac{x}{3}\) = 0

or x(x – \(\frac{1}{3}\)) = 0

or x = 0, \(\frac{1}{3}\)

Question 3.

Check whether x(2x + 3) = x^{2} + 1 is a quadratic equation.

Solution:

x(2x + 3) = x^{2} + 1

⇒ 2x^{2} + 3x = x^{2} + 1

⇒ 2x^{2} + 3x – x^{2} – 1 = 0

⇒ x^{2} + 3x – 1 = 0

Hence, the above equation is a quadratic equation

Question 4.

If one root of the equation x^{2} – 8x + a = 0 is 5, then write the other root.

Solution:

Here the sum of the roots is 8. Therefore the other root will be 8 – 5 = 3.

Question 5.

The sum of the squares of two consecutive natural numbers is 25. If the smaller number is x then write the equation expressing this fact.

Solution:

x^{2} + (x + 1)^{2} = 25

Question 6.

The sum of a number x and its reciprocal is \(\frac{5}{2}\). Write it in the form of an algebraic equation.

Solution:

x + \(\frac{1}{x}\) = \(\frac{5}{2}\)

Question 7.

A positive number x is 56 less than its square. Write the equation expressing this sentence.

Solution:

x^{2} – 56 = x

or x^{2} – x = 56

∴ x^{2} – x – 56 = 0

Question 8.

Write the values of x satisfying the equation \(\frac{x}{5}\) – \(\frac{5}{x}\) = 0.

Solution:

x = ± 5

Question 9.

If the roots of the equation x^{2} – 8x + k = 0 are equal, then find out the value of k.

Solution:

For roots being equal

b^{2} – 4ac = 0

∴ (- 8)^{2} – 4 × 1 × k = 0

Here a = 1. b = – 8 c = k

64 – 4k = 0

∴ k = \(\frac{64}{4}\) = 16

Question 10.

Find the roots of the equation \(\frac{x}{4}\) – \(\frac{4}{x}\) = 0.

Solution:

\(\frac{x}{4}\) – \(\frac{4}{x}\) = 0

∴ \(\frac{x}{4}\) = \(\frac{4}{x}\)

⇒ x^{2} = 16

∴ x = ±\(\sqrt {16}\) = ±4

Question 11.

Solve the equation x – \(\frac{1}{x}\) = 0.

Solution:

x – \(\frac{1}{x}\) = 0

⇒ x = 0 + \(\frac{1}{x}\) = \(\frac{1}{x}\)

⇒ x = \(\frac{1}{x}\)

∴ x^{2} = 1

x = ±\(\sqrt {1}\) = ±1

Hence, x = 1, – 1

Question 12.

Solve the equation 2x^{2} – 8 = 0.

Solution:

The given equation is 2x^{2} – 8 = 0

⇒ 2 (x^{2} – 4) = 0

⇒ 2 (x + 2) (x – 2) = 0

⇒ x = 2, – 2

Question 13.

Check whether (x – 2)^{2} + 1 = 2x – 3 is a quadratic equation.

Solution:

(x – 2)^{2} + 1 = 2x – 3

⇒ x^{2} – 4x + 4 + 1 = 2x – 3

⇒ x^{2} – 4x + 5 – 2x + 3 = 0

⇒ x^{2} – 6x + 8 = 0

Therefore the above equation is a quadratic equation.

Question 14.

Find the nature of the roots of the equation 2x^{2} + 3x + 4 = 0.

Solution:

a = 2, b = 3, c = 4

∴ Value of the discriminant (D)

= b^{2} – 4ac

= (3)^{2} – 4 × 2 × 4

= 9 – 32

= – 23 < 0

∵ The value of the discriminant is negative, therefore the roots do not exist, i.e., the roots will be imaginary.

Question 15.

Write the quadratic formula.

Solution:

If b^{2} – 4ac ≥ 0, then the roots of the quadratic equation ax^{2} + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\). This formula is called the quadratic formula.

Question 16.

Find the value of k for which x = 2 is a root of the quadratic equation 3x^{2} – kx – 2 = 0.

Solution:

∵ x = 2 is a root of the equation 3x^{2} – kx – 2 = 0,

Therefore putting x = 2 in the equation

3(2)^{2} – k(2) -2 = 0

⇒ 12 – 2k – 2 = 0

⇒ – 2k = – 10

∴ k = 5

Short Answer Type Questions—

Question 1.

Write an equation where roots are 6 and – 1.

Solution:

Sum of the roots = 6 + (- 1) = 5

Product of the roots = 6 × (- 1) = – 6

We know that the required equation is x^{2} – (sum of the roots) × + product of the roots = 0

⇒ x^{2} – 5x – 6 = 0

Question 2.

Solve the equation (3x + 2) (2x + 3) = 6.

Solution:

From the given equation—

(3x + 2) (2x +3) = 6

or 6x^{2} + 9x + 4x + 6 = 6

or 6x^{2} + 13x = 0

or x (6x + 13) = 0

or x = 0

or 6x + 13 = 0 ∴ 6x = – 13

Therefore x = 0, \(-\frac{13}{6}\)

Question 3.

Solve the following quadratic equation :

4x^{2} + 4bx – (a^{2} – b^{2}) = 0

Solution:

The given equation is 4x^{2} + 4bx – (a^{2} – b^{2}) = 0

⇒ 4x^{2} + 4bx + b^{2} – a^{2} = 0

⇒ (2x + b)^{2} – a^{2} = 0

⇒ (2x + b + a)(2x + b – a) = 0 [∵ a^{2} – b^{2} = (a + b)(a – b)]

Either 2x + b + a = 0 ⇒ x = \(\frac{-a-b}{2}\) = \(̄\left(\frac{a+b}{2}\right)\)

or 2x + b – a = 0 ⇒ x = \(̄\frac{a-b}{2}\)

Hence the roots of the given equation are

x = –\(-\left(\frac{a+b}{2}\right)\), x = \(\left(\frac{a-b}{2}\right)\)

Question 4.

Solve the equation x + \(\frac{1}{x}\) = 3\(\frac{1}{3}\)

Solution:

The given equation is

⇒ 3(x^{2} + 1) = 10x

⇒ 3x^{2} + 3 = 10x

∴ 3x^{2} – 10x + 3 = 0

or 3x^{2} – 9x – x + 3 = 0

or 3x(x – 3) – 1(x – 3) = 0

or (3x – 1) (x – 3) = 0

∴ x = 3, \(\frac{1}{3}\)

Question 5.

The difference in the ages of Ankit and his father is 30 years. The difference of the squares of their ages is 1560. Find the ages of Ankit and his father.

Solution:

Let the age of father = x years

Then, the age of Ankit = (x – 30) years

According to the question

x^{2} – (x – 30)^{2} = 1560

⇒ x^{2} – x^{2} + 60x – 900 = 1560

⇒ 60x = 1560 + 900

⇒ 60x = 2460

⇒ x = \(\frac{2460}{60}\) = 41

Hence age of father = 41 years and the age of son = 41 – 30 = 11 years

Question 6.

Find the roots of the quadratic equation 6x^{2} – x – 2 = 0.

Solution:

According to the question

6x^{2} – x – 2 = 6x^{2} + 3x – 4x – 2

= 3x(2x + 1) – 2(2x + 1)

= (3x – 2)(2x + 1)

The roots of 6x^{2} – x – 2 = 0 are these values of x for which (3x – 2) (2x + 1) is zero. Therefore,

3x – 2 = 0 or 2x + 1 = 0

i.e., x = \(\frac{2}{3}\) or x = \(-\frac{1}{2}\)

Therefore the roots of 6x^{2} – x – 2 = 0 are \(\frac{2}{3}\) and \(-\frac{1}{2}\).

For the verification of the roots we check whether \(\frac{2}{3}\) and \(-\frac{1}{2}\) satisfy the equation 6x^{2} – x – 2 = 0 or not.

Question 7.

Find the roots of the quadratic equation 3x^{2} – 2\(\sqrt {6}\) x + 2 = 0.

Solution:

Now for x = \(\sqrt{\frac{2}{3}}\), \(\sqrt{3} x-\sqrt{2}\) = 0

Hence this root, on account of the factor \(\sqrt{3} x-\sqrt{2}\) appearing two times, appears two times, i.e., this root is repeated.

Therefore, the roots of 3x^{2} – 2\(\sqrt {6}\) x + 2 = 0 are \(\sqrt{\frac{2}{3}}\), \(\sqrt{\frac{2}{3}}\)

Question 8.

The diagram given below is of a prayer hall. Find the dimensions of this hail.

Solution:

Length of the hall

= (2x + 1) m

Breadth of the hall = x m

∴ Area of the hall =(2x + 1)xm^{2}

= (2x^{2} + x) m^{2}

∴ 2x^{2} + x = 300 (given)

Hence 2x^{2} + x – 300 = 0

Therefore the breadth of the hall must satisfy the equation 2x^{2} + x – 300 = 0 which is a quadratic equation.

Using the factorisation method, we write this equation as follows :

2x^{2} – 24x + 25x – 300 = 0

or 2x(x – 12) + 25(x – 12) = 0

i.e., (x – 12) (2x + 25) = 0

Hence, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, therefore it cannot be negative. Therefore, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m

Question 9.

Solve the equation 2x^{2} – 5x + 3 = 0 by method of completing the square.

Solution:

The equation 2x^{2} – 5x + 3 = 0 is the same as x^{2} – \(\frac{5}{2}\)x + \(\frac{3}{2}\) = 0

or x = \(\frac{3}{2}\) or x = 1

Hence the solutions of the equation are x = \(\frac{3}{2}\) and 1.

Question 10.

Solve the roots of the equation 5x^{2} – 6x – 2 = 0 by the method of completing the square.

Solution:

Multiplying the given equation by 5,

25x^{2} – 30x – 10 = 0

This is equivalent to the following :

(5x)^{2} – 2 × (5x) × 3 + 3^{2} – 3^{2} – 10 = 0

or (5x – 3)^{2} – 9 – 10= 0

or (5x – 3)^{2} – 19=0

or (5x – 3)^{2}= 19

or 5x – 3 = ± \(\sqrt {19}\)

or 5x = 3 ± \(\sqrt {19}\)

or x = \(\frac{3 \pm \sqrt{19}}{5}\)

Hence the roots are \(\frac{3+\sqrt{19}}{5}\) and \(\frac{3-\sqrt{19}}{5}\).

Verify that the roots are \(\frac{3+\sqrt{19}}{5}\) and \(\frac{3-\sqrt{19}}{5}\).

Question 11.

Out of a group of Swans, \(\frac{7}{2}\) times the square root of the total number are playing on the shore of a pond. The two remaining ones are swimming in water. Find the total number of Swans.

Solution:

Let the total number of Swans = x

∴ \(\frac{7}{2}\) times tire square root of x = \(\frac{7}{2}\)\(\sqrt {x}\)

According to the question

\(\frac{7}{2}\)\(\sqrt {x}\) + 2 = x

⇒ \(\frac{7}{2}\)\(\sqrt {x}\) = x – 2

Squaring both sides

\(\frac{49}{4}\)x = x^{2} – 4x + 4

⇒ 49x = 4x^{2} – 16x + 16

⇒ 4x^{2} – 16x + 16 – 49x = 0

⇒ 4x^{2} – 65x +16 = 0

⇒ 4x^{2} – 64x – x + 16 = 0

⇒ 4x(x – 16) – 1(x – 16) = 0

⇒ (4x – 1) (x – 16) = 0

If 4x – 1 = 0

then x = \(\frac{1}{4}\)

If x – 16 = 0

then x = 16

⇒ x = 16, \(\frac{1}{4}\)

Hence the total number of Swans = 16

Long Answer Type Questions—

Question 1.

Two sides of a triangle x cm and 3 (x + 1) cm form a right angle. If the area of the triangle is 84 sq. cm., then find all the sides of the triangle.

Solution:

According to the question

\(\frac{1}{2}\)x × 3 (x+ 1) = 84

( Area of △ = \(\frac{1}{2}\) × Base × Height)

or \(\frac{3 x^{2}+3 x}{2}\) = 84

⇒ 3x^{2} + 3x =168

or 3x^{2} + 3x – 168 = 0

⇒ 3 (x^{2} + x – 56) = 0

∴ x^{2} + x – 56 = \(\frac{0}{3}\) = 0

Hence x^{2} + 8x – 7x – 56 = 0

or x (x + 8) – 7 (x + 8) = 0

or (x + 8) (x – 7) = 0

If x – 7 = 0 then x = 7

If x + 8 = 0

Then x = – 8 which is not possible since the length of the side of the triangle cannot be negative.

Therefore,

One side = 7 cm and second side = 3 (7 + 1) = 24 cm

and third side = \(\sqrt{7^{2}+24^{2}}\)

= \(\sqrt{625}\) = 25 cm.

Question 2.

The sum of two numbers is 48 and their product is 432. Find the numbers.

Solution:

Let the first number = x

∴ second number = (48 – x)

According to the question

x(48 – x) = 432

⇒ 48x – x^{2} = 432

⇒ x^{2} – 48x + 432 = 0

⇒ x^{2} – 36x – 12x + 432 = 0

⇒ x (x – 36) – 12(x – 36) = 0

⇒ (x – 36) (x – 12) = 0

If x – 36 = 0 then x = 36

If x – 12 = 0 then x = 12

Hence the numbers will be 12, 36

Question 3.

₹ 6500 were divided equally among a certain number of persons. Had there been 15 more persons, each would have got ₹ 30 less. Find the original number of persons.

Solution:

Let the original number of persons = x

Therefore share of each person = Rs. \(\frac{6500}{x}\)

According to the question

(x + 15)(\(\frac{6500}{x}\) – 30) = 6500

(x + 15) (6500 – 30x) = 6500x

⇒ 6500x – 30x^{2} + 97500 – 450x – 6500x = 0

⇒ -30x^{2} – 450x + 97500 = 0

⇒ -30(x^{2} + 15x – 3250) = 0

⇒ x^{2} + 15x – 3250 = 0

⇒ x^{2} + 65x – 50x – 3250 = 0

⇒ x (x + 65) – 50(x + 65) = 0

⇒ (x + 65) (x – 50) = 0

∴ Either x + 65 = 0 ⇒ x = – 65 (neglecting)

or x – 50 = 0 ⇒ x = 50

Hence the original number of persons = 50

Question 4.

The length of the hypotenuse of a right triangle is 3\(\sqrt {10}\) cm. If the smaller side is tripled and the larger side is doubled, then the length of the new hypotenuse becomes 9\(\sqrt {5}\) cm. Find all the three sides of the triangles.

Solution:

Let the smaller side = x cm

Then by Pythogoras Theorem the larger side

= \(\sqrt{(3 \sqrt{10})^{2}-x^{2}}\)

= \(\sqrt{90-x^{2}}\) cm

According to the question

(3x)^{2} + (2\(\sqrt{90-x^{2}}\))^{2} = (9\(\sqrt{5}\))^{2}

⇒ 9x^{2} + 4 (90 – x^{2}) = 81 × 5

⇒ 9x^{2} + 360 – 4x^{2} = 405

⇒ 5x^{2} = 405 – 360

⇒ 5x^{2} = 45

⇒ x^{2} = \(\frac{45}{5}\) = 9

⇒ x = ± 3

∵ Side cannot be negative, therefore x = + 3

Hence the smaller side = 3 cm

and the larger side = \(\sqrt{90-(3)^{2}}\)

= \(\sqrt{90-9}\)

= \(\sqrt{81}\) = ± 9

Side cannot be negative, therefore the larger side = 9 cm

Hence the smaller side = 3 cm and the larger side = 9 cm.

Question 5.

The area of a rectangular plot is 528 m^{2}. The length of the region (in metres) is one more than double of its breadth. Solve it by quadratic formula.

Solution:

Let the breadth of the plot be x m.

Then, length = (2x + 1) m

We are given that

x (2x + 1) = 528

i.e., 2x^{2}+ x – 528 = 0

It is of the form ax^{2} + bx + c = 0, where a. = 2, b = 1, c = – 528

Therefore by quadratic formula, we obtain the following solutions

i.e., x = 16 or x = \(\frac{-33}{2}\)

Since x being a dimension cannot be negative. Therefore the breadth of the plot is 16m and the legnth is (2 × 16 + 1) = 33m

Question 6.

Find two consecutive odd positive integers the sum of whose squares is 290.

Solution:

Let the smaller integer out of the two consecutive odd positive integers be x. Then the other integer will be x + 2. According to the question

x^{2} + (x + 2)^{2} = 290

i.e., x^{2} + x^{2} + 4x + 4 = 290

i.e., 2x^{2} + 4x – 286 = 0

i.e., x^{2} + 2x – 143 = 0,

Which is a quadratic equation is x. Comparing this quadratic equation with ax^{2} + bx + c = 0

a = 1, b = 2 and c = – 143

Using the quadratic formula

i.e., x = 11 or x = – 13

But x is given to be an odd positive integer. So x = 11, since x ≠ – 13.

Hence, the two consecutive odd positive integers are 11 and 13.

Question 7.

A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig.)- Find its length and breadth.

Solution:

Let the breadth of the rectangular park be x m.

So, its length = (x + 3) m

Therefore, the area of the rectangular park

= x (x + 3) m^{2}

= (x^{2} + 3x) m^{2}

Now, base of the isosceles triangle = x m

Therefore, its area = \(\frac{1}{2}\) × x × 12

= 6x m^{2}

According to the question

x^{2} + 3x = 6x + 4

or x^{2} – 3x – 4 = 0

Comparing the above quadratic equation with quadratic equation ax^{2} + bx + c.

a = 1, b = – 3 and c = – 4

Using the quadratic formula, we get

But x ≠ 1, So, x = 4 [Length can not be negature]

Therefore,

breadth of the park = 4 m and length of the park = 7 m.

Question 8.

Find the roots of the following quadratic equations, if they exist, using the quadratic formula :

(i) 3x^{2} – 5x + 2 = 0

(ii) x^{2} + 4x + 5 = 0

(iii) 2x^{2} – 2\(\sqrt{2}\)x + 1 = 0

Solution:

(i) For 3x^{2} – 5x + 2

Here a = 3, b = – 5, c = 2

So, b^{2} – 4ac = 25 – 24 = 1 > 0

Therefore, x = \(\frac{5 \pm \sqrt{1}}{6}\) = \(\frac{5 \pm 1}{6}\)

i.e ., x = 1 or x = \(\frac{2}{3}\)

So, the roots are \(\frac{2}{3}\) and 1.

(ii) For x^{2} + 4x + 5 = 0 :

Here a = 1, b = 4, c = 5

So, b^{2} – 4ac = 16 – 20 = -4 < 0

But, since the square of a real number cannot be negative, therefore \(\sqrt{b^{2}-4 a c}\) will not have any real value.

So, there are no real roots for the given equation.

(iii) For2x^{2} – 2\(\sqrt{2}\)x + 1 = 0:

Here a = 2, b = – 2\(\sqrt{2}\), c = 1

So, b^{2} – 4ac = 8 – 8 = 0

Question 9.

Find the roots of the following equations :

(i) x + \(\frac{1}{x}\) = 3, x ≠ 0

(ii) \(\frac{1}{x}\) – \(\frac{1}{x-2}\) = 3, x ≠ 0, 2

Solution:

(i) For x + \(\frac{1}{x}\)

multiplying all the terms by x ≠ 0,

x^{2} + 1 = 3x

⇒ x^{2} – 3x + 1 = 0, which is a quadratic equation.

Here a = 1, b = – 3, c = 1

So, b^{2} – 4ac = 9 – 4 = 5 > 0

(ii) \(\frac{1}{x}\) – \(\frac{1}{x-2}\) = 3, x ≠ 0, 2:

∵ x ≠ 0, 2, therefore multiplying the equation by x (x – 2).

(x – 2) – x = 3x (x – 2)

= 3x^{2} – 6x

So, the given equation after transformation becomes 3x^{2} – 6x + 2 = 0. which is a quadratic equation.

Here, a = 3, b = – 6, c = + 2

So, b^{2} – 4ac = 36 – 24 = 12 > 0

Question 10.

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

Let the speed of the stream be x km/h

Therefore, the speed of the boat upstream = (18 – x) km/h

and the speed of the boat downstream = (18 + x) km/h

The time taken to go upstream = \(\frac{Distance}{Speed}\) = \(\frac{24}{18-x}\) hours

Similarly, the time taken to go downstream = \(\frac{24}{18+x}\) hours

According to the question

\(\frac{24}{18-x}\) – \(\frac{24}{18+x}\) = 1

or 24 (18 + x) – 24(18 – x) = (18 – x) (18 + x)

or 24 × 18 + 24x- 24 × 18 + 24x = (18)^{2} – (x)^{2}

or 48x = 324 – x^{2}

or x^{2} + 48x – 324 = 0

Comparing the above equation with quadratic equation

ax^{2} + bx + c = 0

a = 1, b = 48 and c = – 324

Using the quadratic formula,

Since x is the speed of the stream, therefore it cannot be negative. So, we ignore the root x = – 54. Therefore, from x = 6 we obtain that the speed of the stream is 6 km/ h

Question 11.

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution:

According to the diagram, let the required location of the pole be P. The distance of the pole from the gate B is x m, i.e., BP = .v m. Now the difference of the distances of the pole from the two getes = AP – BP (or BP – AP) = 7 m. Therefore, AP = (x + 7) m

Also. AB = 13 m

Since AB is a diameter.

Therefore, ∠APB = 90°

Therefore AP^{2} + PB^{2} = AB^{2} (By Pythagoras Theorem)

i.e., (x + 7)^{2} + x^{2} = 132

i.e., x^{2} + 14x + 49 + x^{2} = 169

i.e.. 2x^{2} + 14x – 120 = 0

or x^{2} + 7x – 60 = 0

So, the distance ‘x’ of the pole from pole B satisfies the equation x^{2} + 7x – 60 = 0.

Therefore, b^{2} – 4ac = 7^{2} – 4 × 1 × (- 60)

= 289 > 0

Hence. the given quadratic equation has two real roots and therefore it is possible to erect the pole on the boundary of the park.

Solving the quadratic equation x^{2} + 7x – 60 = 0, by the quadratic formula.

Comparing the quadratic equation x^{2} + 7x – 60 = 0 with ax^{2} + bx + c = 0.

a = 1, b = 7 and c = – 60

Therefore, x = 5 or – 12

Since .r is the distance between the pole and the gate B, it must be positive.

Therefore, x = – 12 will have to be ignored. So, x = 5

Thus, the pole has to be eracted on the boundary of the park at a distance of 5 m from the gate B and \(\sqrt{13^{2}-5^{2}}\) = 12 m from the gate A.

Question 12.

An express train takes 2 hours less than a passenger train to travel 400 km between two stations (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 10 km/h more than that of the passenger train, find the average speed of the two trains.

Solution:

Let the speed of the passenger train be x km/h. Then the speed of the express train = (x + 10) km/h, since the speed of the express train is 10 km/h more than the speed of the passenger train.

Now according to the question

or 2x(x + 10) = 4000

or x^{2} + 10x – 2000 = 0

x^{2} + 50x – 40x – 2000 = 0

x(x + 50) – 40(x + 50) = 0

(x + 50) (x – 40) = 0

∴ x = 40, – 50

∵ Speed cannot be negative, therefore

x = 40 km/h

and x + 10 = 40 + 10 = 50 km/h

i.e., the average speed of the passenger train is 40 km/h and the average speed of the express train is 50 km/h.

Question 13.

The speed of a boat in still water is 18 km/h. It takes \(\frac{1}{2}\) an hour extra in going 12 km upstream instead of going the same distance downstream. Find the speed of the stream.

Solution:

Let the speed of the stream be x km/h. Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat downstream = (18 + x) km/h.

The time taken to go upstream

= \(\frac{Distance}{Speed}\) = \(\frac{12}{18-x}\) hour

Similarly, the time taken to go downstream

= \(\frac{12}{18+x}\) hour

According to the question

or 24(18 + x) – 24(18 – x) =(18 + x) (18 – x)

or 24 × 18 + 24x – 24 × 18 + 24x = (18)^{2} – (x)^{2}

or 48x = 324 – x^{2}

or x^{2} + 48x – 324 = 0

Comparing the above equation with quadratic equation

ax^{2} + bx + c = 0

a = 1, b – 48 and c = – 324

Using the quadratic formula

or x = 6 or -54

Since x is the speed of the stream, therefore it cannot be negative. So, we ignore the root x = -54. Therefore, from x = 6, we obtain that the speed of the stream is 6 km/h.

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