Ex.4.2 Q1 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

Find the roots of the following quadratic equations by factorization:

(i) \(\begin{align}\quad x{}^\text{2}-3x-10=0~\end{align}\)

(ii) \(\begin{align}\quad2x{}^\text{2}+x-6=0\end{align}\)

(iii) \(\begin{align}\quad\sqrt {\left( 2 \right)} {x^2} + 7x + 5\sqrt 2 \end{align}\)

(iv) \(\begin{align}\quad2{x^2} - x + 1/8\end{align}\)

(v) \(\begin{align}\quad100x{}^\text{2}-20x+1\end{align}\)

 Video Solution
Quadratic Equations
Ex 4.2 | Question 1

Text Solution

Reasoning:

Roots of the polynomial are same as the zeros of the polynomial. Therefore, roots can be found by factorizing the quadratic equation into two linear factors and equating each to zero.

What is Known?

Quadratic Equations. 

What is unknown?

Roots of given Quadratic Equations.

(i) \(\begin{align}\quad x{}^\text{2}-3x-10=0~\end{align}\)

Steps:

\[\begin{align} x - 5x + 2x - 10 &= 0 \\ x \left( {x - 5} \right) + 2\left( {x - 5} \right) & = 0 \\ \left( {x - 5} \right)\left( {x + 2} \right) & = 0 \end{align}\]

\[\begin{align} x-5 & = 0 \quad \; \rm{and}& x+ 2 &= 0 \\  x &=5 \quad &x &= - 2 \\\end{align} \]

Roots are \(– 2,\; 5.\)

(ii) \(\begin{align}\quad2x{}^\text{2}+x-6=0\end{align}\)

Steps:

\[\begin{align} 2 x^{2}+x-6 &=0 \\ 2 x^{2}+4 x-3 x-6 &=0 \\ 2 x(x+2)-3(x+2) &=0 \\(2 x-3)(x+2) &=0  \end{align}\]

\[\begin{align}   2 x-3 &=0 \qquad \text{and} \qquad x+2=0\\ 2 x &=3 \qquad \text{and} \quad \qquad x=-2 \\ x &=3 / 2\qquad \text{and} \qquad x=-2 \end{align}\]

Roots are: \(\begin{align}3/2, - 2\,\end{align}\) 

(iii) \(\begin{align}\quad\sqrt {\left( 2 \right)} {x^2} + 7x + 5\sqrt 2 \end{align}\)

Steps:

\[\begin{align}\sqrt {\left( 2 \right)} x + 7x + 5\sqrt 2 &= 0\\\sqrt {\left( 2 \right)} x + 5x + 2x + 5\sqrt 2 &= 0\\ \begin{bmatrix} \sqrt {\left( 2 \right)} x + 5x \\ + \sqrt {\left( 2 \right)} \times \sqrt {\left( 2 \right)} x \\ + 5\sqrt 2 \end{bmatrix} &= 0\\ \begin{bmatrix} x\left( {\sqrt {\left( 2 \right)} x + 5} \right) \\ + \sqrt 2 \left( {\sqrt {\left( 2 \right)} x + 5} \right) \end{bmatrix} &= 0\\\left( {x\sqrt {\left( 2 \right)} + 5} \right)\left( {x + \sqrt 2 } \right) &= 0 \end{align}\]

\[ \begin{align} &\left( {\sqrt {\left( 2 \right)} x + 5} \right) = 0\quad  \left( {x + \sqrt 2 } \right) = 0\\ & \sqrt {\left( 2 \right)} x = - 5 \quad \qquad x = - \sqrt 2 \\ & x =\!\! \left( \frac{ - 5}{ \sqrt {\left( 2 \right)}} \right) \qquad x = - \sqrt 2 \end{align}\]

Roots are \(\begin{align}\frac{ - 5}{ \sqrt {\left( 2 \right)}} , - \sqrt 2 .\end{align}\)

(iv) \(\begin{align}\quad2{x^2} - x + 1/8\end{align}\)

Steps:

Multiplying both sides of the equation by \(8\)

\[\begin{align} 2(8) x^{2}-8(x)+(8) 1 / 8 &=0(8) \\ 16 x^{2}-8 x+1 &=0 \\ 16 x^{2}-4 x-4 x+1 &=0 \\ 4 x(4 x-1)-1(4 x-1) &=0 \\(4 x-1)-1(4 x-1) &=0 \end{align}\]

\[\begin{align}  4 x-1 &=0 \qquad \qquad 4 x-1 =0 \\ 4 x &=1 \!\qquad \qquad \qquad 4 x =1 \\ x &=1 / 4 \qquad \qquad \quad \; x =1 / 4\end{align}\]

Roots are \(\begin{align}~1/4\,\,,\,\,1/4~.\end{align}\)

(v) \(\begin{align}\quad100x{}^\text{2}-20x+1\end{align}\)

Steps:

\[\begin{align} 100 x^{2}-20 x+1 &=0 \\ 100 x^{2}-10 x-10 x+1 &=0 \\ 10 x(10 x-1)-1(10 x-1) &=0 \\ (10 x-1)(10 x-1) &=0 \end{align} \]

\[\begin{align} 10 x-1 &=0 \qquad \qquad10 x-1=0 \\ 10 x &=1\qquad \qquad\qquad \!10 x=1 \\ x &=\frac{1}{10} \qquad \qquad\quad x=\frac{1 }{10} \end{align}\]

Roots are \(\begin{align} \frac{1}{10},\, \frac{1}{10} \,\,.\end{align}\)

  
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