Question

The Positive root of √5x²+√2x = 4 is​

Answer 1

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     5*x^2-(2*x+4)=0 

Step by step solution :

STEP1:Equation at the end of step 1

5x2 - (2x + 4) = 0

STEP2:Trying to factor by splitting the middle term

 2.1     Factoring  5x2-2x-4 

The first term is,  5x2  its coefficient is  5 .

The middle term is,  -2x  its coefficient is  -2 .

The last term, "the constant", is  -4 

Step-1 : Multiply the coefficient of the first term by the constant   5 • -4 = -20 

Step-2 : Find two factors of  -20  whose sum equals the coefficient of the middle term, which is   -2 .

     -20   +   1   =   -19     -10   +   2   =   -8     -5   +   4   =   -1     -4   +   5   =   1     -2   +   10   =   8     -1   +   20   =   19

Answer 2

Step-by-step explanation:

$\sqrt{5} {x}^{2} + \sqrt{2} x - 4 = 0 \\ a = \sqrt{5} \: \: , \: \: b = \sqrt{2} \: \: , \: \: c = - 4\\ D = {b}^{2} - 4ac \\ = {( \sqrt{2} )}^{2} - 4 \times \sqrt{5} \times ( - 4) \\ = 2 + 16 \sqrt{5} \\ x = \frac{ - b± \sqrt{D} }{2a} \\ = \frac{ - b+ \sqrt{D} }{2a} \: \: , \: \: \frac{ - b- \sqrt{D} }{2a} \\ = \frac{ - \sqrt{2} + \sqrt{2 + 16 \sqrt{5} } }{2 \times \sqrt{5} } \: \: , \: \: \frac{ - \sqrt{2} - \sqrt{2 + 16 \sqrt{5} } }{2 \times \sqrt{5} } \\ = \frac{ - \sqrt{2} + \sqrt{2 } \times \sqrt{1 + 8 \sqrt{5} } }{ \sqrt{2} \times \sqrt{2} \times \sqrt{5} } \: \: , \: \: \frac{ - \sqrt{2} - \sqrt{2} \times \sqrt{1 + 8\sqrt{5} } }{ \sqrt{2} \times \sqrt{2} \times \sqrt{5} } \\ = \frac{ - 1+ \sqrt{1 + 8\sqrt{5} } }{ \sqrt{10} } \times \frac{ - \sqrt{10} }{ \sqrt{10} } \: \: , \: \: \frac{ - 1 - \sqrt{1 + 8 \sqrt{5} } }{ \sqrt{10} } \times \frac{ - \sqrt{10} }{ \sqrt{10} } \\ = \frac{ \sqrt{10} - \sqrt{10 + 80 \sqrt{5} } }{10 } \: \: , \: \: \frac{ \sqrt{10} + \sqrt{10 + 80 \sqrt{5} } }{10 } \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$