Question

criterion of equation x2-√7x-4=0 is answer 32,23,19??​

Answer 1

➡ USER ANSWER IS :-

➡ 32,23,19

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GIVEN :-

➡ x2-√7x-4=0

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SOLUTION :-

$\large \red\leadsto \red{x \times 2 - \sqrt{7x} - 4 = 0}$

$\large \red\leadsto \red{2x - \sqrt{7x} - 4 = 0}$

$\large \red\leadsto \red{ - \sqrt{7x} = - 2x + 4}$

$\large \red\leadsto \red{ \sqrt{7x} = 2x - 4}$

$\large \red\leadsto \red{7x = 4x {}^{2} - 16x + 16}$

$\large \red\leadsto \red{23x - 4x {}^{2} - 16 = 0}$

$\large \red\leadsto \red{- 4x {}^{2} + 23x - 16 = 0}$

$\large \red\leadsto \red{4x {}^{2} - 23x + 16 = 0}$

$\large \red\leadsto \red{x = \frac{ - ( - 23) \frac{ + }{} \sqrt{( - 23) {}^{2} - 4 \times 4 \times 16} }{2 \times 4} }$

$\large \red\leadsto \red{x = \frac{23 \frac{ + }{} \sqrt{529 - 256} }{8} }$

$\large \red\leadsto \red{x = \frac{23 \frac{ + }{} \sqrt{273} }{8} }$

CHECKING THE SOLUTION :-

$\large \red\leadsto \red{x = \frac{23 + \sqrt{273} }{8} }$

$\large \red\leadsto \red{ \frac{23 - \sqrt{273} }{8}}$

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SIMPLIFY :-

$1) \tiny \red\leadsto \red{ \frac{23 + \sqrt{273} }{8} \times 2 - \sqrt{7 \times \frac{23 + \sqrt{273} }{8} - 4 = 0} }$

$2)\tiny \red\leadsto \red{ \frac{23 - \sqrt{273} }{8} \times 2 - \sqrt{7 \times \frac{23 - \sqrt{273} }{8} - 4 = 0 } }$

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WHICH IS SOLUTION AND WHICH IS NOT :-

$1)\red\leadsto \blue{0 = 0}$

$2)\red\leadsto \blue{ - 4.76136 = 0}$

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SO THE SOLUTION IS :-

$\huge \red\leadsto \red{x = \frac{23 + \sqrt{273} }{8}}$

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ALTERNATE FORM :-

$\huge \red\leadsto \bold\red{x = 4.94034}$

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SO YOUR ANSWER 32,23,19 is wrong :)

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