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# Ex.4.3 Q2 Quadratic Equations Solutions - NCERT Maths Class 10

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## Question

Find the roots of the quadratic equations given in Q1 above by applying quadratic formula.

Video Solution
Ex 4.3 | Question 2

## Text Solution

What is known?

What is Unknown?

Reasoning:

If the given quadratic equation is: $$ax{}^{2}+bx+c=0$$, then:

If $$b^{2}-4ac\ge 0$$ then the roots are \begin{align} x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \end{align}

If $$b{}^{2}-4ac<0$$ then no real roots exist.

Steps:

i)

\begin{align}a &= 2,\;b = - 7,\;c = 3\\{b^2} - 4ac &= {\left( { - 7} \right)^2} - 4\left( 2 \right)\left( 3 \right)\\&= 49 - 24\\{b^2} - 4ac &= 25 > 0\end{align}

The Roots are,

\begin{align} x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ x &= \frac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(2)(3)} }}{{2(2)}}\\x &= \frac{{ - ( - 7) \pm \sqrt {49 - 24} }}{{2(2)}}\\x &= \frac{{7 \pm 5}}{4}\\x &= \frac{{7 + 5}}{4}\qquad x = \frac{{7 - 5}}{4}\\x &= \frac{{12}}{4}\qquad \quad x = \frac{2}{4}\\x &= 3 \qquad \qquad x = \frac{1}{2}\end{align}

The Roots are,

\begin{align}3,\frac{1}{2} \end{align}

ii)

\begin{align}{{a}} &= 2,\;{{b}} = 1,\;{{c}} = - 4\\{{{b}}^2} - 4{{ac}} &= {{(1)}^2} - 4(2)( - 4)\\ &= 1 + 32\\& = 33 > 0\end{align}

The Roots are,

\begin{align} x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{{2(2)}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{4}\\x& = \frac{{ - 1 + \sqrt {33} }}{4}\\x &= \frac{{ - 1 - \sqrt {33} }}{4}\end{align}

The Roots are,

\begin{align}\frac{{ - 1 + \sqrt {33} }}{4},\;\frac{{ - 1 - \sqrt {33} }}{4} \end{align}

iii)

\begin{align}a &= 4,b = (4\sqrt 3 ),c = 3\\{b^2} - 4ac& = {(4\sqrt 3 )^2} - 4(4)(3)\\ &= (16 \times 3) - (16 \times 3)\\{b^2} - 4ac &= 0\end{align}

The Roots are,

\begin{align}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - b \pm 0}}{{2a}}\\&= \frac{{ - b}}{{2a}}\\&= \frac{{ - 4\sqrt 3 }}{{2(4)}}\\x &= \frac{{ - \sqrt 3 }}{2}\end{align}

The Roots are,

\begin{align}\frac{{ - \sqrt 3 }}{2},\;\frac{{ - \sqrt 3 }}{2}.\end{align}

iv)

\begin{align}a &= 2, b = 1, c = 4\\{b^2} - 4ac &= {{(1)}^2} - 4(2)(4)\\ &= 1 - 32\\&= - 31 < 0\\{b^2} - 4ac &< 0\end{align}

$$\therefore$$ No real roots are exist.

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