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X-3÷x+3-x+3÷x-3=48÷7 |
| Answer» {tex}\\Rightarrow{/tex}{tex}(x - 3)(x - 3) - (x + 3)(x + 3) = \\frac{48}{7}(x + 3)(x - 3){/tex} [Multiplying both sides by (x + 3)(x -3)]{tex}\\Rightarrow{/tex}{tex}x^2 + 9 - 6x - [x^2 + 9 + 6x] ={/tex}{tex}\\frac{{48}}{7}{/tex}{tex}[(x^2 - 9)]{/tex}{tex}\\Rightarrow{/tex}{tex}x^2 + 9 - 6x - x^2 - 9 - 6x ={/tex}{tex}\\frac{{48}}{7}{/tex}{tex}(x^2 - 9){/tex}{tex}\\Rightarrow - 12x = \\frac{{48}}{7}({x^2} - 9){/tex}{tex} \\Rightarrow - 12x \\times 7 = 48({x^2} - 9){/tex} [Multiplying both sides by 7]{tex}\\Rightarrow - 7x = \\frac{{48}}{{12}}({x^2} - 9){/tex} [Dividing both sides by 12]{tex}\\Rightarrow{/tex}{tex}-7x = 4(x^2 - 9){/tex}{tex}\\Rightarrow{/tex}{tex} -7x = 4x^2 - 36{/tex}{tex}\\Rightarrow{/tex}{tex}4x^2 + 7x - 36 = 0{/tex}In order to factorize 4x2 + 7x - 36, we have to find two numbers \'a\' and \'b\' such that.a + b = 7 and a {tex}\\times {/tex} b = 4 {tex}\\times{/tex} (-36) = -144Clearly, 16 + (-9) = 7 and (16) {tex}\\times{/tex} (-9) = -144{tex}\\therefore{/tex} a = 16 and b = -9Now,4x2 + 7x - 36 = 0{tex}\\Rightarrow{/tex} 4x2 + 16x - 9x - 36 = 0{tex}\\Rightarrow{/tex} 4x(x + 4) - 9(x + 4) = 0{tex}\\Rightarrow{/tex} (x + 4)(4x - 9) = 0{tex}\\Rightarrow{/tex} x + 4 = 0 or 4x - 9 = 0{tex}\\Rightarrow{/tex} x = -4 or {tex}x = \\frac{9}{4}{/tex}Hence, the required roots of given equation are -4 and\xa0{tex}\\frac{{9}}{4}{/tex} | |