MATH SOLVE

2 months ago

Q:
# Find the value of the lesser root of x^2-6x+8=0

Accepted Solution

A:

x² - 6x +8= 0,

[tex]x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} a=1, b=-6, c=8 [/tex]

[tex]x= \frac{6+/- \sqrt{36-4*1*8} }{2*1} x=(6+/- \sqrt{36-32} )/2 x=(6+/-2)/2 x_{1} =4, x_{2}=2[/tex]

[tex]x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} a=1, b=-6, c=8 [/tex]

[tex]x= \frac{6+/- \sqrt{36-4*1*8} }{2*1} x=(6+/- \sqrt{36-32} )/2 x=(6+/-2)/2 x_{1} =4, x_{2}=2[/tex]