Ex.4.4 Q1 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

Find the nature of the roots of the following quadratic equations, if the real root exists, find them.

(i) \(2x^2-3x+5=0\)

(ii) \(3x^2 - 4\sqrt {\left( 3 \right)} x + 4 = 0\)

(iii) \(2x^{2}-6x+3=0\)

 Video Solution
Quadratic Equations
Ex 4.4 | Question 1

Text Solution

What is Unknown?

Nature of roots.

Reasoning:

The general form of a quadratic equation is \(ax^{2}+bx+c=0\)

\(b^{2}{ }-{ }4ac\) is called the discriminant of the quadratic equation and we can decide whether the real roots are exist or not based on the value of the discriminant:

i) Two distinct real roots, if \(b^{2}{ }-{ }4ac{ }>{ }0\)

ii) Two equal real roots, if \(b^{2}{ }-{ }4ac{ }={ }0\)

iii) No real roots, if \(b^{2}{ }-{ }4ac{ }<{ }0\)

(i) \(2x^2-3x+5=0\)

Steps:

(i) \(2x^2-3x+5=0\)

\(a = 2, \;b = -3\; c = 5\)

\[\begin{align}{b^2} - 4ac &= {{( - 3)}^2} - 4(2)(5)\\&= 9 - 40\\&= - 31|\\{b^2} - 4ac &> 0\end{align}\]

\(\therefore \) No real roots.

(ii) \(3x^2 - 4\sqrt {\left( 3 \right)} x + 4 = 0\)

\(a = 3,\;b = - 4\sqrt ( 3),\;c = 4\)

\[\begin{align}{b^2} - 4ac &= {\left( { - 4\sqrt {(3} )} \right)^2} - 4(3)(4)\\&= 16 \times 3 - 4 \times 4 \times 3\\&= (16 \times 3) - (16 \times 3)\\&= 0\\{b^2} - 4ac &= 0\end{align}\]

\(\therefore\;\)Two equal real roots.

\[\begin{align}{\text{ Sum of roots }} &= 2x{ = - \frac{b}{a}}\\x& = - \frac{b}{{2a}}\\x &= \frac{{ - ( - 4\sqrt 3 )}}{{2(3)}}\\&= \frac{{4\sqrt 3 }}{{2\sqrt 3 \sqrt 3 }}\\&= \frac{2}{{\sqrt 3 }}\end{align}\]

Roots are \(\begin{align} \frac{2}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}. \end{align}\)

(iii) \(2x^{2}-6x+3=0\)

\(a = 2,\; b = -6, \;c = 3\)

\[\begin{align}{b^2} - 4ac &= {{( - 6)}^2} - 4(2)(3)\\&= 36 - 24\\&= 12|\\{b^2} - 4ac &> 0\end{align}\]

\(\therefore\;\)Two distinct real roots.

\[\begin{align}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
x &= \frac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)(3)} }}{{2(2)}}\\&= \frac{{6 \pm \sqrt {12} }}{4}\\
x &= \frac{{6 \pm 2\sqrt 3 }}{4}\\x &= \frac{{3 \pm \sqrt 3 }}{2}\end{align}\]

Roots are \(\begin{align}x = \frac{{3 + \sqrt 3 }}{2},\frac{{3 - \sqrt 3 }}{2}.\end{align}\)

  
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