(a+b)²x²-4abx-(a-b)²=0 Find by factarisation
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Sia ? 6 years, 6 months ago
We have,
{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}
In order to factorize {tex}(a + b)^2x^2 - 4abx - (a - b)^2{/tex}, we have to find two numbers 'l' and 'm' such that.
l + m = -4ab and lm = -(a + b)2(a - b)2
Clearly, (a - b)2 + [-(a + b)2] = -4ab and lm = -(a + b)2(a - b)2
l = (a - b)2 and m = -(a + b)2
Now,
{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(a + b)^2x^2 - (a + b)^2x + (a - b)^2x - (a - b)^2 = 0{/tex}
{tex}\Rightarrow{/tex} (a + b)2x [x - 1] + (a - b)2[x - 1] = 0
{tex}\Rightarrow{/tex} (x - 1)[(a + b)2x + (a - b)2] = 0
{tex}\Rightarrow{/tex} x - 1 = 0 or (a + b)2x + (a - b)2 = 0
{tex}\Rightarrow{/tex} x = 1 or {tex}x = - \frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}}{/tex}
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