Ex.4.1 Q1 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

Check whether the followings are quadratic equations:

i)   \(\begin{align}{(x + 1)^2} = 2\left( {x - 3} \right)\end{align}\)

ii)  \(\begin{align} \left( {{x^2} - 2x} \right) = \left( { - 2} \right)\left( {3 - x} \right)\end{align}\)

iii) \(\begin{align} \begin{bmatrix}\left( {x - 2} \right)\left( {x + 1} \right) \\ = \left( {x - 1} \right)\left( {x + 3} \right) \end{bmatrix}\end{align}\)

iv) \(\begin{align} \left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)\end{align}\)

v)  \(\begin{align} \begin{bmatrix} \left( {2x - 1} \right)\left( {x - 3} \right) \\ = \left( {x + 5} \right)\left( {x - 1} \right) \end{bmatrix} \end{align}\)

vi) \(\begin{align} {x^2} + 3x + 1 = {\left( {x - 2} \right)^2}\end{align}\)

vii)  \(\begin{align} {\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)\end{align}\)

viii) \(\begin{align} {x^3} - 4x - x + 1 = {\left( {x - 2} \right)^3}\end{align}\)

 Video Solution
Quadratic Equations
Ex 4.1 | Question 1

Text Solution

Reasoning:

Standard form of a quadratic equation is \(\begin{align}a{x^2} + bx + c = 0\end{align}\) where \(\begin{align}a \ne 0\end{align}\) and degree of quadratic equation is \(2.\)

What is unknown?

To check whether the given equation is a quadratic equation

Steps:

i)

\(\begin{align}\quad{(x + 1)^2} = 2\left( {x - 3} \right)\end{align}\)

\[\begin{align}&{\left( {{{a + b}}} \right)^{{2}}} = {{{a}}^{{2}}}{{ + }}{{{b}}^{{2}}}{{ + 2ab}}\\&{{{x}}^{{2}}}{{ + 2x + 1}} = \left( {{{2x - 6}}} \right)\\&{{{x}}^{{2}}}{{ + 2x + 1 - }}\left( {{{2x - 6}}} \right) = 0\\&x^2+2x+1-2x+6\\&{{{x}}^{{2}}}{{ + 7}} = 0\end{align}\]

Here degree of \(\begin{align}x^2 + 7 = 0\end{align}\)is \(2.\)

\(\therefore\)  It is a quadratic equation.

ii)

\(\begin{align}\quad\left( {{x^2} - 2x} \right) = \left( { - 2} \right)\left( {3 - x} \right)\end{align}\)

\[\begin{align}{x^2} - 2x &= - 6 + 2x\\{x^2}\, - 2x - 2x + 6 &= 0\\{x^2} - 4x + 6 &= 0\end{align}\]

Degree \(= 2.\)

\(\therefore\)  It is a quadratic equation.

iii)

\(\begin{align}\quad\left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)\end{align}\)

\[\begin{align}\begin{bmatrix}{x^2} - 2x + \\x - 2 \end{bmatrix}&=   {x^2} + 3x  - x - 3  \\{x^2} - x - 2 &= {x^2} + 2x - 3\\\begin{bmatrix}{x^2} - x - 2 - \\{x^2} - 2x + 3\end{bmatrix} &= 0\\ - 3x + 1 &= 0\end{align}\]

Degree \(= 1\)

\(\therefore\) It is not a quadratic equation.

iv)

\(\begin{align}\quad\left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)\end{align}\)

\[\begin{align}&2{x^2} + x - 6x - 3 = {x^2} + 5x\\&2{x^2} - 5x - 3 = {x^2} + 5x\\&2{x^2} - 5x - 3 - {x^2} - 5x = 0\\&{x^2} - 10x - 3 = 0\end{align}\]

Degree \(= 2\)

\(\therefore\) It is a quadratic equation.

v)

\(\begin{align}\quad\left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)\end{align}\)

\[\begin{align}\begin{bmatrix}2{x^2} \! - \! 6x \! -\\ x \! + \! 3\end{bmatrix} & \! = \! {x^2} \! - \! x \! + \! 5x \! - \! 5  \\2{x^2} \! - \! 7x \! + \! 3 & \! = \! {x^2} \! + \! 4x \! - \! 5\\\begin{bmatrix}2{x^2} \! - \! 7x \! + \! 3 \! -\\ {x^2} \! - \! 4x \! + \! 5 \end{bmatrix}& \! = \! 0\\{x^2} \! - \! 11x \! + \! 8 & \! = \! 0\end{align}\]

Degree \(= 2\)

\(\therefore\) It is a quadratic equation

vi)

\(\begin{align}{}\quad{x^2} + 3x + 1 = {\left( {x - 2} \right)^2}\end{align}\)

\[\begin{align} {x^2} + 3x + 1 &= {x^2} - 4x + 4 \\   [\because ( a - b )^2 &= {a^2} - 2ab + {b^2} ] \\ \begin{bmatrix} {x^2} + 3x + 1 - \\{x^2} + 4x - 4\end{bmatrix} &= 0 \\ 7x - 3 &= 0\end{align}\]

Degree \(= 1\)

\(\therefore\) It is not a quadratic equation.

vii)

\(\begin{align}\quad{\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)\end{align}\)

Since,

\[\begin{align}&{\left( {a + b} \right)^3} =  {a^3} + 3{a^2}b+   3a{b^2} + {b^3}  \end{align}\]

\[\begin{align} & \begin{bmatrix} {x^3} + 3{x^2}\left( 2 \right) +\\ 3{\left( 2 \right)^2}x + {\left( 2 \right)^3} \end{bmatrix}= 2{x^3} - 2x \\& \begin{bmatrix} {x^3} + 6{x^2} + 12x +\\ 8 - 2{x^3} + 2x \end{bmatrix} = 0\\& - {x^3} + 6{x^2} + 14x + 8 = 0\end{align}\]

Degree \(= 3\)

\(\therefore\) It is not a quadratic equation.

viii)

\(\begin{align}\quad{x^3} - 4x - x + 1 = {\left( {x - 2} \right)^3}\end{align}\)

\[\begin{align} \begin{bmatrix} {x^3} \! - \! 4{x^2} \! - \! \\ x \! + \! 1 \end{bmatrix} & \! = \! \begin{bmatrix} {x^3} \! - \! 3 {\left( x \right)^2}\left( 2 \right) \! + \! \\ 3\left( x \right){\left( 2 \right)^2} \! - \! {\left( 2 \right)^3} \end{bmatrix} \\ \because \, \left( a \! - \! b \right)^3 & \! = \! {a^3} \! - \! 3{a^2}b \! + \! 3a{b^2} \! - \! {b^3} \\ \begin{bmatrix} {x^3} \! - \! 4{x^2} \! -\\ x \! + \! 1\end{bmatrix} & \! = \! \begin{bmatrix} {x^3} \! - \! 6{x^2} \! + \! \\ 12x \! - \! 8 \end{bmatrix} \\ \begin{bmatrix}{x^3} \! - \! 4{x^2} \! - \! x \! +\\ 1 \! - \! {x^3} \! + \! 6{x^2} \! - \! \\ 12x \! + \! 8 \end{bmatrix} & \! = \! 0\\ 2{x^2} \! - \! 13x \! + \! 9 & \! = \! 0 \end{align}\]

Degree \(= 2\)

\(\therefore\) It is a quadratic equation

  
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