Saved Bookmarks
| 1. |
Find the root by factorization 2x²—x+1/8=0 |
| Answer» We have, {tex}2x^2-x+{1\\over8}=0{/tex}{tex}\\implies 2x^2-{1\\over2}x-{1\\over2}x+{1\\over8}=0{/tex}{tex}\\implies x(2x-{1\\over2}) - {1\\over4}(2x-{1\\over2})=0{/tex}{tex}\\implies (2x-{1\\over2}) (x-{1\\over4})=0{/tex}{tex}Either \\,(2x-{1\\over2})\\, =0\\,or\\, (x-{1\\over4})=0{/tex}{tex}\\implies x = {1\\over4},\\,{1\\over4}{/tex}So, this root is repeated root.{tex}\\therefore {/tex} both the roots are {tex}{1\\over4}{/tex}. | |