NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 4: Quadatric Equation

Exercise 4.4

General idea: For a quadratic equation of the form \(ax^2 + bx + c = 0\), we use the discriminant \(D = b^2 - 4ac\). - If \(D > 0\), we get two real and different roots. - If \(D = 0\), we get two equal real roots. - If \(D < 0\), we get no real roots.

(i) \(8x^2 + 2x - 3 = 0\)

Here, \(a = 8, b = 2, c = -3\).

Step 1: Find discriminant. \(D = b^2 - 4ac = 2^2 - 4(8)(-3) = 4 + 96 = 100\).

Step 2: Since \(D > 0\), there are 2 real and different roots.

Step 3: Use formula \(x = \dfrac{-b \pm \sqrt{D}}{2a}\).

\(x = \dfrac{-2 \pm 10}{16}\).

So, \(x = -\tfrac{3}{4}\) or \(x = \tfrac{1}{2}\).

(ii) \(-2x^2 + 3x + 2 = 0\)

First, multiply the whole equation by \(-1\) to make it easier: \(2x^2 - 3x - 2 = 0\).

Here, \(a = 2, b = -3, c = -2\).

Step 1: \(D = (-3)^2 - 4(2)(-2) = 9 + 16 = 25\).

Step 2: \(D > 0\), so two real and different roots.

Step 3: Formula: \(x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{3 \pm 5}{4}\).

So, \(x = -\tfrac{1}{2}\) or \(x = 2\).

(iii) \(5x^2 - 2x - 10 = 0\)

Here, \(a = 5, b = -2, c = -10\).

Step 1: \(D = (-2)^2 - 4(5)(-10) = 4 + 200 = 204\).

Step 2: \(D > 0\), so two real and different roots.

Step 3: \(\sqrt{204} = 2\sqrt{51}\).

Step 4: Formula: \(x = \dfrac{2 \pm 2\sqrt{51}}{10} = \dfrac{1 \pm \sqrt{51}}{5}\).

(iv) \(\dfrac{1}{2x-3} + \dfrac{1}{x-5} = 1\)

Step 1: Take LCM on the left side:

\(\dfrac{(x-5) + (2x-3)}{(2x-3)(x-5)} = 1\).

So, numerator: \(3x - 8\).

Step 2: Cross multiply: \(3x - 8 = (2x-3)(x-5)\).

Expand: \(2x^2 - 10x - 3x + 15 = 2x^2 - 13x + 15\).

So the equation becomes: \(2x^2 - 16x + 23 = 0\).

Step 3: Find discriminant: \(D = (-16)^2 - 4(2)(23) = 256 - 184 = 72\).

\(\sqrt{72} = 6\sqrt{2}\).

Step 4: Formula: \(x = \dfrac{16 \pm 6\sqrt{2}}{4} = 4 \pm \tfrac{3\sqrt{2}}{2}\).

(v) \(x^2 + 5\sqrt{5}x - 70 = 0\)

Here, \(a = 1, b = 5\sqrt{5}, c = -70\).

Step 1: \(D = (5\sqrt{5})^2 - 4(1)(-70) = 125 + 280 = 405\).

Step 2: \(D > 0\), so real and different roots.

Step 3: \(\sqrt{405} = 9\sqrt{5}\).

Step 4: Formula: \(x = \dfrac{-5\sqrt{5} \pm 9\sqrt{5}}{2}\).

Simplify: roots are \(-7\sqrt{5}\) and \(2\sqrt{5}\).

Step 1: Suppose the required natural number is \(x\).

Step 2: The question says "the square diminished by 84". That means: \(x^2 - 84\).

Step 3: It also says "is equal to thrice of (8 more than the number)". That means: \(3(x + 8)\).

Step 4: Write the equation: \(x^2 - 84 = 3(x + 8)\).

Step 5: Expand the right-hand side: \(x^2 - 84 = 3x + 24\).

Step 6: Bring everything to one side: \(x^2 - 3x - 108 = 0\).

Step 7: Factorise this quadratic equation: \(x^2 - 3x - 108 = (x - 12)(x + 9)\).

Step 8: From this, either \(x - 12 = 0\) → \(x = 12\), or \(x + 9 = 0\) → \(x = -9\).

Step 9: Since a natural number must be positive, we take \(x = 12\).

Final Answer: The number is 12.

Step 1: Let the natural number be \(x\). Since it is a natural number, \(x > 0\).

Step 2: The reciprocal of \(x\) is \(\dfrac{1}{x}\). The question says: "When the number is increased by 12, it equals 160 times its reciprocal." So, the equation is:

\(x + 12 = 160 \times \dfrac{1}{x}\)

or, \(x + 12 = \dfrac{160}{x}\).

Step 3: Remove the fraction by cross multiplying:

\(x(x + 12) = 160\)

\(x^2 + 12x = 160\)

\(x^2 + 12x - 160 = 0\).

Step 4: This is a quadratic equation. Solve it using the quadratic formula:

Formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(ax^2 + bx + c = 0\).

Here, \(a = 1, b = 12, c = -160\).

Step 5: Substitute values:

\(x = \dfrac{-12 \pm \sqrt{12^2 - 4(1)(-160)}}{2}\)

\(x = \dfrac{-12 \pm \sqrt{144 + 640}}{2}\)

\(x = \dfrac{-12 \pm \sqrt{784}}{2}\)

\(x = \dfrac{-12 \pm 28}{2}\).

Step 6: Now calculate both values:

• \(x = \dfrac{-12 + 28}{2} = \dfrac{16}{2} = 8\)
• \(x = \dfrac{-12 - 28}{2} = \dfrac{-40}{2} = -20\)

Step 7: Since \(x\) must be a natural number (positive), we take \(x = 8\).

Final Answer: The required natural number is 8.

Step 1: Assume the original speed

Let the original speed of the train be \(v\,\text{km/h}\).

Step 2: Write the time formula

Time taken to travel a distance = \(\dfrac{\text{distance}}{\text{speed}}\).

Step 3: Time at the original speed

Time = \(\dfrac{360}{v}\) hours.

Step 4: Time at the increased speed

If the speed is increased by 5 km/h, new speed = \(v + 5\).

Time = \(\dfrac{360}{v+5}\) hours.

Step 5: Difference in times

It is given that the difference in times is 48 minutes.

Convert 48 minutes to hours: \(48\,\text{minutes} = \dfrac{48}{60} = \dfrac{4}{5}\,\text{hours}\).

Step 6: Form the equation

\(\dfrac{360}{v} - \dfrac{360}{v+5} = \dfrac{4}{5}\).

Step 7: Simplify

Take LCM of denominators: \(v(v+5)\).

\(\dfrac{360(v+5) - 360v}{v(v+5)} = \dfrac{4}{5}\).

\(\dfrac{1800}{v(v+5)} = \dfrac{4}{5}\).

Step 8: Cross multiply

\(1800 \times 5 = 4 \times v(v+5)\).

\(9000 = 4v^2 + 20v\).

Step 9: Simplify equation

Divide everything by 2: \(4500 = 2v^2 + 10v\).

Rearrange: \(2v^2 + 10v - 4500 = 0\).

Divide by 2: \(v^2 + 5v - 2250 = 0\).

Step 10: Solve quadratic equation

Factorise: \(v^2 + 5v - 2250 = 0\).

\((v + 50)(v - 45) = 0\).

So, \(v = -50\) or \(v = 45\).

Step 11: Choose positive value

Since speed cannot be negative, the answer is \(v = 45\).

Final Answer: The original speed of the train is 45 km/h.

Step 1: Assume her present age.

Let Zeba's present age be \(x\) years.

Step 2: Write the condition given in the question.

If Zeba were 5 years younger, her age would be \(x - 5\).

The square of this age is \((x - 5)^2\).

This square is equal to "five times her actual age + 11".

So, the equation is: \((x - 5)^2 = 5x + 11\).

Step 3: Expand the square.

\((x - 5)^2 = x^2 - 10x + 25\).

So the equation becomes: \(x^2 - 10x + 25 = 5x + 11\).

Step 4: Bring all terms to one side.

\(x^2 - 10x + 25 - 5x - 11 = 0\).

\(x^2 - 15x + 14 = 0\).

Step 5: Solve the quadratic equation.

Factorise: \(x^2 - 15x + 14 = (x - 14)(x - 1) = 0\).

So, \(x = 14\) or \(x = 1\).

Step 6: Check which answer is possible.

If \(x = 1\), then her age 5 years ago would be negative, which is not possible.

So, the only valid answer is \(x = 14\).

Final Answer: Zeba’s age is 14 years.

Step 1: Suppose Nisha’s present age is \(n\) years.

Step 2: We are told Asha’s present age is 2 more than the square of Nisha’s age. So Asha’s age = \(n^2 + 2\). Let this be \(a\).

Step 3: The difference in their ages is \(a - n\). This is the number of years after which Nisha will reach her mother’s current age.

Step 4: At that time, Asha’s age will also increase by \(a - n\) years. So Asha’s age will then be \(a + (a - n) = 2a - n\).

Step 5: The question says that when Nisha is as old as her mother is now, Asha’s age will be one less than 10 times Nisha’s current age. So, \(2a - n = 10n - 1\).

Step 6: Substitute \(a = n^2 + 2\) into this equation: \(2(n^2 + 2) - n = 10n - 1\).

Step 7: Simplify: \(2n^2 + 4 - n = 10n - 1\) \(2n^2 - 11n + 5 = 0\).

Step 8: Solve this quadratic equation. Possible values of \(n\) are 5 or 0.5. Since age must be a whole number, \(n = 5\).

Step 9: Now find Asha’s age: \(a = n^2 + 2 = 5^2 + 2 = 27\).

Final Answer: Nisha is 5 years old and Asha is 27 years old.

Step 1: The whole lawn is a rectangle with size \(50 \times 40\).

Total area of lawn = \(50 \times 40 = 2000\,\text{m}^2\).

Step 2: A pond is made in the centre. The grass is the part outside the pond.

Grass area is given as \(1184\,\text{m}^2\).

Step 3: This means:

Grass area = Total lawn area – Pond area

\(1184 = 2000 - \text{Pond area}\)

So, Pond area = \(2000 - 1184 = 816\,\text{m}^2\).

Step 4: Now, let the width of the grass border (the part outside pond) be \(x\) metres.

Then, length of pond = \(50 - 2x\) (because \(x\) is cut from both left and right).

Breadth of pond = \(40 - 2x\) (because \(x\) is cut from top and bottom).

Step 5: Pond area = \((50 - 2x)(40 - 2x)\).

We know Pond area = 816, so:

\((50 - 2x)(40 - 2x) = 816\).

Step 6: Expand:

\(2000 - 100x - 80x + 4x^2 = 816\).

\(2000 - 180x + 4x^2 = 816\).

\(4x^2 - 180x + 1184 = 0\).

Divide by 4: \(x^2 - 45x + 296 = 0\).

Step 7: Solve the quadratic:

\(x^2 - 45x + 296 = 0\).

By factorisation: \((x - 8)(x - 37) = 0\).

So, \(x = 8\) or \(x = 37\).

Step 8: Can \(x = 37\)?

No, because if border = 37, then pond size becomes negative. So only \(x = 8\) is correct.

Step 9: Pond dimensions:

Length = \(50 - 2(8) = 34\,\text{m}\).

Breadth = \(40 - 2(8) = 24\,\text{m}\).

Final Answer: Length = 34 m, Breadth = 24 m.

Step 1: At 2 pm, the minute hand is at 12. At exactly 3 pm, the minute hand will come back to 12 again. So from 2 pm to 3 pm, there are 60 minutes in total.

Step 2: If it is already \(t\) minutes past 2 pm, then the time still left to reach 3 pm is \(60 - t\) minutes.

Step 3: According to the question, this remaining time is 3 minutes less than \(\dfrac{t^2}{4}\). So we can write the equation: \[ 60 - t = \dfrac{t^2}{4} - 3 \]

Step 4: Remove the fraction by multiplying everything by 4: \[ 4(60 - t) = t^2 - 12 \] Simplify: \[ 240 - 4t = t^2 - 12 \]

Step 5: Bring all terms to one side: \[ t^2 + 4t - 252 = 0 \]

Step 6: Factorise the quadratic: \[ (t - 6)(t + 42) = 0 \]

Step 7: So \(t = 6\) or \(t = -42\). But \(t\) represents minutes past 2 pm, so it must be between 0 and 60. Therefore, \(t = 6\).

Final Answer: The correct time is 6 minutes past 2 pm.