Question

help!!!!!!!
Find the roots of the above quadratic equations ( if they exist) by using quadratic formula.
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Answer 1

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(ii) x² - 2ax +(a² - b²) =0

or, x² - 2ax + (a +b) (a-b) =0

or, x² - {(a+b) +(a-b)}x + (a+b) (a-b) =0

or, x² - (a+b)x - (a-b)x + (a +b) (a-b) =0

or, x{x-(a+b)} - (a-b) {x-(a+b)} =0

or, (x- a-b) (x - a +b) =0

So... The possible roots are..
x = a + b
and
x = a - b.
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(iii)
$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4} \\ or. \: \frac{(x + 2) + 2(x + 1)}{(x + 2)(x + 1)} = \frac{4}{x + 4} \\ or. \: \frac{3x + 4}{( {x}^{2} + 3x + 2) } = \frac{4}{x + 4} \\ or. \: (3x + 4)(x + 4) = 4( {x}^{2} + 3x + 2) \\ or. \: 3 {x}^{2} + 4x + 12x + 16 = 4 {x}^{2} + 12x + 8 \\ or. \: {x}^{2} - 4x - 8 = 0 \\ \\ by \: shreedhar \: acharyas \: formula \: \: \\ x = \frac{ - ( - 4) + - \sqrt{ {( - 4)}^{2} - 4 \times 1 \times ( - 8)} }{2 \times 1 } \\ = (4 + - \sqrt{16 + 32} ) \div 2 \\ = \frac{4 + - \sqrt{48} }{2} = \frac{4 + - (6.92)}{2} \\ so ...either \: \\ x = (4 + 6.92) \div 2 = \frac{10.92}{2} = 5.46 \\ \\ or..x \: = (4 - 6.92) \div 2 = - \frac{2.92}{2} = - 1.46$



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