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

Go back to  'Ex.4.3'

Question

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

 Video Solution
Quadratic Equations
Ex 4.3 | Question 2

Text Solution

What is known?

A quadratic equation.

What is Unknown?

Roots of the quadratic equation.

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)

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

\(\therefore\;\) Roots are \(\begin{align} x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \end{align}\)

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

\(\therefore \;\)Roots are \(\begin{align}3,\frac{1}{2} \end{align}\)

ii)

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

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

\(\therefore\;\)Roots are, \(\begin{align}\frac{{ - 1 + \sqrt {33} }}{4},\;\frac{{ - 1 - \sqrt {33} }}{4} \end{align}\)

iii)

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

\[\begin{align}\therefore \quad {\rm{ Roots\, are \;}}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
 &= \frac{{ - b \pm 0}}{{2a}}\\&= \frac{{ - b}}{{2a}}\\&= \frac{{ - 4\sqrt 3 }}{{2(4)}}\\x &= \frac{{ - \sqrt 3 }}{2}\end{align}\]

\(\therefore\;\)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.