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The sum of a number and it\'s positive square root is 6\\25 find the ne |
| Answer» Let the required numbers be x and its square root is y. Then,{tex}x + y = \\frac{6}{{25}}{/tex}{tex} \\Rightarrow {y^2} + y = \\frac{6}{{25}}\\;\\left[ {\\because \\sqrt x = y \\Rightarrow x = {y^2}} \\right]{/tex}{tex} \\Rightarrow {/tex}{tex}25y^2 + 25y = 6{/tex} [Multiplying both sides of equation by 25]{tex} \\Rightarrow {/tex}{tex}25y^2 + 25y - 6 = 0{/tex}{tex} \\Rightarrow {/tex}{tex}25y^2 + 30y - 5y - 6 = 0{/tex}{tex} \\Rightarrow {/tex}{tex}5y(5y + 6) - 1(5y + 6) = 0{/tex}{tex} \\Rightarrow {/tex} (5y + 6)(5y - 1) = 0{tex} \\Rightarrow {/tex} 5y - 1 = 0 [{tex}\\because{/tex}{tex}\\sqrt x{/tex} > 0 {tex}\\Rightarrow{/tex} y > 0 {tex} \\Rightarrow {/tex} 5y + 6 {tex}\\ne{/tex} 0]{tex} \\Rightarrow y = \\frac{1}{5}{/tex}{tex} \\Rightarrow \\sqrt x = \\frac{1}{5}{/tex}{tex} \\Rightarrow x = \\frac{1}{{25}}{/tex} [Taking square on both sides]Hence, required number is {tex}\\frac{1}{{25}}{/tex}. | |