Answer:
The value of lesser root is:
x=2
Step-by-step explanation:
We are given a quadratic equation in terms of variable " x " as:
[tex]x^2-6x+8=0[/tex]
We know that for any quadratic equation of the type:
[tex]ax^2+bx+c=0[/tex]
The roots of x are calculated as:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here we have:
a=1 , b=-6 and c=8
Hence, on solving for roots:
[tex]x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4\times 1\times 8}}{2\times 1}\\\\\\x=\dfrac{6\pm \sqrt{36-32}}{2}\\\\\\x=\dfrac{6\pm \sqrt{4}}{2}\\\\\\x=\dfrac{6\pm 2}{2}[/tex]
Hence, we have:
[tex]x=\dfrac{6+2}{2}\ or\ x=\dfrac{6-2}{2}\\\\x=\dfrac{8}{2}\ or\ x=\dfrac{4}{2}\\\\\\x=4\ or\ x=2[/tex]
Hence, the value of lesser root is:
x=2
