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

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Question

Check whether the followings are quadratic equations:

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

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

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

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

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

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

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

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

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

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

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

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

Degree \(= 2\)

\(\therefore\) It is a quadratic equation

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

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

Degree \(= 1\)

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

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

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

Degree \(= 3\)

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

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

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

Degree \(= 2\)

\(\therefore\) It is a quadratic equation

  
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