Quadratic Equations — NCERT Solutions

Given: $2x^2 - x + \dfrac{1}{8} = 0$

Multiplying throughout by 8 to clear the fraction:
$$16x^2 - 8x + 1 = 0$$

Splitting the middle term: We need two numbers whose product is $16 \times 1 = 16$ and sum is $-8$. These are $-4$ and $-4$.

$$16x^2 - 4x - 4x + 1 = 0$$
$$4x(4x - 1) - 1(4x - 1) = 0$$
$$(4x - 1)(4x - 1) = 0$$
$$(4x - 1)^2 = 0$$

Setting the factor to zero:
$$4x - 1 = 0 \implies x = \frac{1}{4}$$

The equation has two equal roots: $x = \dfrac{1}{4}$ and $x = \dfrac{1}{4}$.

Given: $100x^2 - 20x + 1 = 0$

Splitting the middle term: We need two numbers whose product is $100 \times 1 = 100$ and sum is $-20$. These are $-10$ and $-10$.

$$100x^2 - 10x - 10x + 1 = 0$$
$$10x(10x - 1) - 1(10x - 1) = 0$$
$$(10x - 1)(10x - 1) = 0$$
$$(10x - 1)^2 = 0$$

Setting the factor to zero:
$$10x - 1 = 0 \implies x = \frac{1}{10}$$

The equation has two equal roots: $x = \dfrac{1}{10}$ and $x = \dfrac{1}{10}$.

Problem (a): Rectangular plot (from Exercise 4.1, Q2(i))

We formed the equation: $2x^2 + x - 528 = 0$, where $x$ = breadth.

Splitting the middle term: Product $= 2 \times (-528) = -1056$, sum $= 1$. Numbers: $33$ and $-32$.

$$2x^2 + 33x - 32x - 528 = 0$$
$$x(2x + 33) - 16(2x + 33) = 0$$
$$(2x + 33)(x - 16) = 0$$

$$x = 16 \quad \text{or} \quad x = -\frac{33}{2}$$

Since breadth cannot be negative, $x = 16\,\text{m}$.

$$\text{Breadth} = 16\,\text{m}, \quad \text{Length} = 2(16) + 1 = 33\,\text{m}$$

Problem (b): Two consecutive integers (from Exercise 4.1, Q2(ii))

We formed the equation: $x^2 + x - 306 = 0$.

Splitting the middle term: Product $= -306$, sum $= 1$. Numbers: $18$ and $-17$.

$$x^2 + 18x - 17x - 306 = 0$$
$$x(x + 18) - 17(x + 18) = 0$$
$$(x + 18)(x - 17) = 0$$

$$x = 17 \quad \text{or} \quad x = -18$$

Since the integers are positive, $x = 17$.

The two consecutive positive integers are $17$ and $18$.

Let the two numbers be $x$ and $27 - x$.

Given: Their product $= 182$
$$x(27 - x) = 182$$
$$27x - x^2 = 182$$
$$x^2 - 27x + 182 = 0$$

Splitting the middle term: Product $= 182$, sum $= -27$. Numbers: $-13$ and $-14$.

$$x^2 - 13x - 14x + 182 = 0$$
$$x(x - 13) - 14(x - 13) = 0$$
$$(x - 13)(x - 14) = 0$$

$$x = 13 \quad \text{or} \quad x = 14$$

If $x = 13$: the other number $= 27 - 13 = 14$.
If $x = 14$: the other number $= 27 - 14 = 13$.

The two numbers are $13$ and $14$.

Let the two consecutive positive integers be $x$ and $x + 1$.

Given: $x^2 + (x+1)^2 = 365$

$$x^2 + x^2 + 2x + 1 = 365$$
$$2x^2 + 2x + 1 - 365 = 0$$
$$2x^2 + 2x - 364 = 0$$
$$x^2 + x - 182 = 0$$

Splitting the middle term: Product $= -182$, sum $= 1$. Numbers: $14$ and $-13$.

$$x^2 + 14x - 13x - 182 = 0$$
$$x(x + 14) - 13(x + 14) = 0$$
$$(x + 14)(x - 13) = 0$$

$$x = 13 \quad \text{or} \quad x = -14$$

Since $x$ is a positive integer, $x = 13$.

The two consecutive positive integers are $13$ and $14$.

Verification: $13^2 + 14^2 = 169 + 196 = 365$ ✓

Let the base of the right triangle $= x\,\text{cm}$.

Then the altitude $= (x - 7)\,\text{cm}$.

Using Pythagoras' theorem:
$$\text{base}^2 + \text{altitude}^2 = \text{hypotenuse}^2$$
$$x^2 + (x - 7)^2 = 13^2$$
$$x^2 + x^2 - 14x + 49 = 169$$
$$2x^2 - 14x + 49 - 169 = 0$$
$$2x^2 - 14x - 120 = 0$$
$$x^2 - 7x - 60 = 0$$

Splitting the middle term: Product $= -60$, sum $= -7$. Numbers: $-12$ and $5$.

$$x^2 - 12x + 5x - 60 = 0$$
$$x(x - 12) + 5(x - 12) = 0$$
$$(x - 12)(x + 5) = 0$$

$$x = 12 \quad \text{or} \quad x = -5$$

Since the side of a triangle cannot be negative, $x = 12\,\text{cm}$.

$$\text{Base} = 12\,\text{cm}, \quad \text{Altitude} = 12 - 7 = 5\,\text{cm}$$

The base is $12\,\text{cm}$ and the altitude is $5\,\text{cm}$.

Let the number of articles produced in a day $= x$.

Then the cost of production of each article $= ₹(2x + 3)$.

Given: Total cost $= ₹90$
$$x(2x + 3) = 90$$
$$2x^2 + 3x = 90$$
$$2x^2 + 3x - 90 = 0$$

Splitting the middle term: Product $= 2 \times (-90) = -180$, sum $= 3$. Numbers: $15$ and $-12$.

$$2x^2 + 15x - 12x - 90 = 0$$
$$x(2x + 15) - 6(2x + 15) = 0$$
$$(2x + 15)(x - 6) = 0$$

$$x = 6 \quad \text{or} \quad x = -\frac{15}{2}$$

Since the number of articles cannot be negative or a fraction, $x = 6$.

$$\text{Number of articles} = 6$$
$$\text{Cost of each article} = 2(6) + 3 = ₹15$$

The number of articles produced is $6$ and the cost of each article is ₹$15$.

Given: $2x^2 - 3x + 5 = 0$

Here $a = 2,\ b = -3,\ c = 5$.

Discriminant:
$$D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$$

Since $D = -31 < 0$,

The equation has no real roots.

Given: $3x^2 - 4\sqrt{3}\,x + 4 = 0$

Here $a = 3,\ b = -4\sqrt{3},\ c = 4$.

Discriminant:
$$D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$$

Since $D = 0$, the equation has two equal (coincident) real roots.

Finding the roots:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{4\sqrt{3} \pm 0}{2 \times 3} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$$

The two equal roots are $x = \dfrac{2\sqrt{3}}{3}$ and $x = \dfrac{2\sqrt{3}}{3}$.

Given: $2x^2 - 6x + 3 = 0$

Here $a = 2,\ b = -6,\ c = 3$.

Discriminant:
$$D = b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12$$

Since $D = 12 > 0$, the equation has two distinct real roots.

Finding the roots:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{12}}{2 \times 2} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$$

$$x_1 = \frac{3 + \sqrt{3}}{2}, \qquad x_2 = \frac{3 - \sqrt{3}}{2}$$

The two distinct real roots are $x = \dfrac{3 + \sqrt{3}}{2}$ and $x = \dfrac{3 - \sqrt{3}}{2}$.

Given: $2x^2 + kx + 3 = 0$

Here $a = 2,\ b = k,\ c = 3$.

Condition for two equal roots: $D = 0$
$$b^2 - 4ac = 0$$
$$k^2 - 4(2)(3) = 0$$
$$k^2 - 24 = 0$$
$$k^2 = 24$$
$$k = \pm\sqrt{24} = \pm 2\sqrt{6}$$

The values of $k$ are $k = 2\sqrt{6}$ or $k = -2\sqrt{6}$.

Given: $kx(x - 2) + 6 = 0$

Expanding:
$$kx^2 - 2kx + 6 = 0$$

Here $a = k,\ b = -2k,\ c = 6$.

Condition for two equal roots: $D = 0$
$$b^2 - 4ac = 0$$
$$(-2k)^2 - 4(k)(6) = 0$$
$$4k^2 - 24k = 0$$
$$4k(k - 6) = 0$$
$$k = 0 \quad \text{or} \quad k = 6$$

Since $k = 0$ would make the equation non-quadratic (coefficient of $x^2$ becomes 0), we reject $k = 0$.

The value of $k$ is $k = 6$.

Let the breadth of the rectangular mango grove $= x\,\text{m}$.

Then the length $= 2x\,\text{m}$.

Given: Area $= 800\,\text{m}^2$
$$\text{Length} \times \text{Breadth} = 800$$
$$2x \times x = 800$$
$$2x^2 = 800$$
$$x^2 = 400$$
$$x^2 - 400 = 0$$

Here $a = 1,\ b = 0,\ c = -400$.

Discriminant:
$$D = b^2 - 4ac = 0 - 4(1)(-400) = 1600 > 0$$

Since $D > 0$, the equation has two distinct real roots, so it is possible to design such a grove.

Finding $x$:
$$x^2 = 400 \implies x = \pm 20$$

Since breadth cannot be negative, $x = 20\,\text{m}$.

$$\text{Breadth} = 20\,\text{m}, \quad \text{Length} = 2 \times 20 = 40\,\text{m}$$

Yes, it is possible. The breadth is $20\,\text{m}$ and the length is $40\,\text{m}$.

Let the present age of one friend $= x$ years.

Then the present age of the other friend $= (20 - x)$ years.

Four years ago:

• Age of first friend $= (x - 4)$ years
• Age of second friend $= (20 - x - 4) = (16 - x)$ years

Given: Product of their ages four years ago $= 48$
$$(x - 4)(16 - x) = 48$$
$$16x - x^2 - 64 + 4x = 48$$
$$-x^2 + 20x - 64 = 48$$
$$-x^2 + 20x - 64 - 48 = 0$$
$$-x^2 + 20x - 112 = 0$$
$$x^2 - 20x + 112 = 0$$

Here $a = 1,\ b = -20,\ c = 112$.

Discriminant:
$$D = b^2 - 4ac = (-20)^2 - 4(1)(112) = 400 - 448 = -48$$

Since $D = -48 < 0$, the equation has no real roots.

The given situation is not possible.

Let the length of the rectangular park $= x\,\text{m}$.

Given: Perimeter $= 80\,\text{m}$
$$2(\text{length} + \text{breadth}) = 80$$
$$\text{length} + \text{breadth} = 40$$
$$\text{breadth} = (40 - x)\,\text{m}$$

Given: Area $= 400\,\text{m}^2$
$$x(40 - x) = 400$$
$$40x - x^2 = 400$$
$$x^2 - 40x + 400 = 0$$

Here $a = 1,\ b = -40,\ c = 400$.

Discriminant:
$$D = b^2 - 4ac = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$$

Since $D = 0$, the equation has two equal real roots, so it is possible to design such a park.

Finding $x$:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{40 \pm 0}{2} = 20\,\text{m}$$

$$\text{Length} = 20\,\text{m}, \quad \text{Breadth} = 40 - 20 = 20\,\text{m}$$

Yes, it is possible. The park is a square with length $=$ breadth $= 20\,\text{m}$.