Quadratic Equations

Given: $(x + 1)^2 = 2(x - 3)$

Simplifying LHS and RHS:
$$x^2 + 2x + 1 = 2x - 6$$
$$x^2 + 2x + 1 - 2x + 6 = 0$$
$$x^2 + 7 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 1,\ b = 0,\ c = 7$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

Given: $x^2 - 2x = (-2)(3 - x)$

Simplifying RHS:
$$x^2 - 2x = -6 + 2x$$
$$x^2 - 2x - 2x + 6 = 0$$
$$x^2 - 4x + 6 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 1,\ b = -4,\ c = 6$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

Given: $(x - 2)(x + 1) = (x - 1)(x + 3)$

Expanding LHS:
$$x^2 + x - 2x - 2 = x^2 - x - 2$$

Expanding RHS:
$$x^2 + 3x - x - 3 = x^2 + 2x - 3$$

Setting LHS = RHS:
$$x^2 - x - 2 = x^2 + 2x - 3$$
$$-x - 2x - 2 + 3 = 0$$
$$-3x + 1 = 0$$

This is not of the form $ax^2 + bx + c = 0$ (the $x^2$ terms cancel, so $a = 0$).

Conclusion: The given equation is not a quadratic equation.

Given: $(x - 3)(2x + 1) = x(x + 5)$

Expanding LHS:
$$2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$$

Expanding RHS:
$$x^2 + 5x$$

Setting LHS = RHS:
$$2x^2 - 5x - 3 = x^2 + 5x$$
$$2x^2 - x^2 - 5x - 5x - 3 = 0$$
$$x^2 - 10x - 3 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 1,\ b = -10,\ c = -3$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

Given: $(2x - 1)(x - 3) = (x + 5)(x - 1)$

Expanding LHS:
$$2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$$

Expanding RHS:
$$x^2 - x + 5x - 5 = x^2 + 4x - 5$$

Setting LHS = RHS:
$$2x^2 - 7x + 3 = x^2 + 4x - 5$$
$$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$$
$$x^2 - 11x + 8 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 1,\ b = -11,\ c = 8$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

Given: $x^2 + 3x + 1 = (x - 2)^2$

Expanding RHS:
$$(x-2)^2 = x^2 - 4x + 4$$

Setting LHS = RHS:
$$x^2 + 3x + 1 = x^2 - 4x + 4$$
$$x^2 - x^2 + 3x + 4x + 1 - 4 = 0$$
$$7x - 3 = 0$$

This is not of the form $ax^2 + bx + c = 0$ (the $x^2$ terms cancel).

Conclusion: The given equation is not a quadratic equation.

Given: $(x + 2)^3 = 2x(x^2 - 1)$

Expanding LHS:
$$(x+2)^3 = x^3 + 3(x^2)(2) + 3(x)(4) + 8 = x^3 + 6x^2 + 12x + 8$$

Expanding RHS:
$$2x(x^2 - 1) = 2x^3 - 2x$$

Setting LHS = RHS:
$$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$$
$$x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0$$
$$-x^3 + 6x^2 + 14x + 8 = 0$$

This is a polynomial of degree 3 (cubic equation), not of the form $ax^2 + bx + c = 0$.

Conclusion: The given equation is not a quadratic equation.

Given: $x^3 - 4x^2 - x + 1 = (x - 2)^3$

Expanding RHS:
$$(x-2)^3 = x^3 - 3(x^2)(2) + 3(x)(4) - 8 = x^3 - 6x^2 + 12x - 8$$

Setting LHS = RHS:
$$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$$
$$x^3 - x^3 - 4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$$
$$2x^2 - 13x + 9 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 2,\ b = -13,\ c = 9$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

Let the breadth of the plot $= x$ metres.

Then the length of the plot $= (2x + 1)$ metres.

Given: Area $= 528\,\text{m}^2$

$$\text{Length} \times \text{Breadth} = 528$$
$$(2x + 1) \times x = 528$$
$$2x^2 + x = 528$$
$$2x^2 + x - 528 = 0$$

This is the required quadratic equation, where $x$ represents the breadth of the plot.

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

Given: Their product $= 306$

$$x(x + 1) = 306$$
$$x^2 + x = 306$$
$$x^2 + x - 306 = 0$$

This is the required quadratic equation, where $x$ is the smaller of the two consecutive integers.

Let Rohan's present age $= x$ years.

Then his mother's present age $= (x + 26)$ years.

3 years from now:
- Rohan's age $= (x + 3)$ years
- Mother's age $= (x + 26 + 3) = (x + 29)$ years

Given: Product of their ages 3 years from now $= 360$

$$(x + 3)(x + 29) = 360$$
$$x^2 + 29x + 3x + 87 = 360$$
$$x^2 + 32x + 87 - 360 = 0$$
$$x^2 + 32x - 273 = 0$$

This is the required quadratic equation, where $x$ is Rohan's present age.

Let the speed of the train $= x\,\text{km/h}$.

Time taken at speed $x$:
$$t_1 = \frac{480}{x}\,\text{hours}$$

Time taken at speed $(x - 8)$:
$$t_2 = \frac{480}{x - 8}\,\text{hours}$$

Given: $t_2 - t_1 = 3$

$$\frac{480}{x-8} - \frac{480}{x} = 3$$

$$480\left(\frac{x - (x-8)}{x(x-8)}\right) = 3$$

$$480 \times \frac{8}{x(x-8)} = 3$$

$$\frac{3840}{x(x-8)} = 3$$

$$3840 = 3x(x - 8)$$

$$1280 = x^2 - 8x$$

$$x^2 - 8x - 1280 = 0$$

This is the required quadratic equation, where $x$ is the speed of the train in km/h.

Given: $x^2 - 3x - 10 = 0$

Splitting the middle term: We need two numbers whose product is $-10$ and sum is $-3$. These are $-5$ and $+2$.

$$x^2 - 5x + 2x - 10 = 0$$
$$x(x - 5) + 2(x - 5) = 0$$
$$(x - 5)(x + 2) = 0$$

Setting each factor to zero:
$$x - 5 = 0 \implies x = 5$$
$$x + 2 = 0 \implies x = -2$$

The roots of the equation are $x = 5$ and $x = -2$.

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

Splitting the middle term: We need two numbers whose product is $2 \times (-6) = -12$ and sum is $+1$. These are $+4$ and $-3$.

$$2x^2 + 4x - 3x - 6 = 0$$
$$2x(x + 2) - 3(x + 2) = 0$$
$$(x + 2)(2x - 3) = 0$$

Setting each factor to zero:
$$x + 2 = 0 \implies x = -2$$
$$2x - 3 = 0 \implies x = \frac{3}{2}$$

The roots of the equation are $x = -2$ and $x = \dfrac{3}{2}$.

Given: $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$

Splitting the middle term: We need two numbers whose product is $\sqrt{2} \times 5\sqrt{2} = 10$ and sum is $7$. These are $5$ and $2$.

$$\sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = 0$$
$$x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0$$
$$(\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0$$

Setting each factor to zero:
$$\sqrt{2}\,x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$$
$$x + \sqrt{2} = 0 \implies x = -\sqrt{2}$$

The roots of the equation are $x = -\dfrac{5\sqrt{2}}{2}$ and $x = -\sqrt{2}$.

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, $x = 16\,\text{m}$.

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

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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 $= 20\,\text{m}$.