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| 1. |
Find the zeroes of √3x2-8x+4√3 |
| Answer» {tex}\\sqrt 3x^2-8x+4\\sqrt 3=0{/tex}{tex}=> \\sqrt 3x^2-6x-2x+4\\sqrt 3=0{/tex}{tex}=> \\sqrt 3x(x-2\\sqrt 3)-2(x-2\\sqrt 3)=0{/tex}{tex}=> (\\sqrt 3x-2)(x-2\\sqrt 3)=0 {/tex}{tex}=> (\\sqrt 3x-2)= 0 ,(x-2\\sqrt 3)=0 {/tex}{tex}=> x ={2\\over \\sqrt 3},2\\sqrt 3{/tex} | |