Question

Root5x²-5root6x-root20=0. find x

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\sqrt{5} x ^{2} - 5 \sqrt{6} x - \sqrt{20} = 0$

• a= √5. b = -5√6. c= -√20

$let \: us \: find \: the \: discriminent \: (d)$

• b²-4ac= 0 ....

$= > b ^{2} - 4ac \\ \\ = > ( - 5 \sqrt{6} )^{2} - 4 \times \sqrt{5} \times - \sqrt{20} \\ \\ \\ = > 150 + 400 \\ \\ = > 550$

Now by applying Sridharacharyas formula we can find the value of X

$= > x = \frac{ - b \: plus \: minus \sqrt{b ^{2} - 4ac} }{2a}$

$= > x = \frac{5 \sqrt{6} \: plus \: minus \sqrt{550} }{2 \sqrt{5} }$

$= > the \: roots \: are \: \: \: eiher \: \: negtive \:or \: positiv$

$= > x = \frac{5 \sqrt{6} + \sqrt{550} }{2a} \: \: or \: x = \frac{5 \sqrt{6} - \sqrt{550} }{2a}$