Chapter 4: Quadratic Equations

Ans: We are given an equation: ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

We will simplify the given equation.

${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 2x + 1 = 2x - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 1 = - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 7 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x = - 6 + 2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 4x + 6 =0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

Ans: We are given an equation: $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

We will simplify the given equation.

$\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x + x - 2 = }{{\text{x}}^{\text{2}}}\text{ - x + 3x - 3}$

$\Rightarrow \text{- 2x + x - 2 = - x + 3x - 3}$

$\Rightarrow \text{- x - 2 = 2x - 3}$

$\Rightarrow \text{3x - 1 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

Ans: We are given an equation: $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

We will simplify the given equation.

$\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{- 6x + x - 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{- 10x - 3 = 0}\]

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

Ans: We are given an equation: $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

We will simplify the given equation.

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - x - 6x + 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x - x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 7x + 3 = 4x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 11x + 8 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\text{x}}^{\text{2}}}\text{ - 4x + 4}$

$\Rightarrow \text{3x + 1 = - 4x + 4}$

$\Rightarrow \text{7x - 3 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

Ans: We are given an equation: ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

We will simplify the given equation.

${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ + 6}{{\text{x}}^{\text{2}}}\text{ + 12x + 8 = 2}{{\text{x}}^{\text{3}}}\text{ - 2x}$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}\text{ + 14x + 8 = }{{\text{x}}^{\text{3}}}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ - 14x - 8 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

Ans: We are given an equation: ${{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}$

We will simplify the given equation.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{- 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - 13x + 9 =  0}$

This equation obtained is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

Ans: It is given that the area of rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$.

Let us assume the breadth of the plot be $\text{b = x m}$.

It is given that the length is one more than twice the breadth.

So, the length will be $\text{l = }\left( \text{2x + 1} \right)\text{ m}$.

The area of the rectangle will be  $\text{A = l  }\!\!\times\!\!\text{  b}$

$\therefore \text{l  }\!\!\times\!\!\text{  b = 528}$

$\Rightarrow \text{x}\left( \text{2x + 1} \right)\text{ = 528}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$

Therefore, the required quadratic equation is $\text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$.

Ans: It is given that the product of two consecutive positive integers is $\text{306}$.

If we assume the first integer to be $\text{x}$, then the next consecutive integer will be $\left( \text{x+1} \right)$.

So, we can write $\text{x}\left( \text{x + 1} \right)\text{ = 306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$.

Ans: Let us assume Rohan’s age be $\text{x}$.

His mother is $\text{26}$ years older than him.

So, his mother’s age will be $\left( x+26 \right)$

After $\text{3}$ years, Rohan’s age $\text{= x + 3}$

His mother’s age $\text{= x + 26 + 3 = x + 29}$

It is given that after $\text{3}$ years, the product of their ages is $\text{360}$.

$\therefore \left( \text{x + 3} \right)\left( \text{x + 29} \right)\text{ = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x + 87 = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$.

Ans: Let us assume the speed of the train be $\text{x km/h}$.

The train travels a distance of $\text{480 km}$ at a uniform speed.

So, the time taken to travel the given distance $\text{= }\frac{\text{480}}{\text{x}}\text{ hrs}$.

It is given that the train would take $\text{3}$ hours more if the

speed had been $\text{8 km/h}$ less.

Now, the new speed of the train $\text{= }\left( \text{x - 8} \right)\text{ km/h}$.

The new time taken to travel the same distance $\text{= }\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ hr}$.

We know that $\text{speed  }\!\!\times\!\!\text{  time = distance}$

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

$\Rightarrow \text{480 + 3x - }\frac{\text{3840}}{\text{x}}\text{ - 24 = 480}$

$\Rightarrow \text{3x - }\frac{\text{3840}}{\text{x}}\text{ = 24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{ - 24x - 3840 = 0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$.