Question

Brian is solving the equation 2x-3/4x=5 What value must be added to both sides of the equation to make the left side a perfect-square trinomial?

Answer 1

yess u bring to 2 as a fraction then put 5 above a x

Answer 2

Answer:

$(2\sqrt{2}x-\frac{5}{\sqrt{2}})^{2}=\frac{31}{2}}$

Step by step solution

$2x-\frac{3}{4x}=5$

Multiplying each term with $4x$ we get

$2x*4x-3=5*4x$

$8x^2-3=20x$

$8x^2-20x-3=0$

Splitting the above expression in the form of

$a^2-2*a*b+b^2$

where $a=2\sqrt{2}$

$((2\sqrt{2}*x)^{2})-2*(2\sqrt{2})*(\frac{5}{\sqrt{2}})-3$

adding and subtracting

$(\frac{5}{\sqrt{2}})^{2}$

to the above polynomial..

$(2\sqrt{2}*x)^{2}-2*2\sqrt{2}*(\frac{5}{\sqrt{2}})+(\frac{5}{\sqrt{2}})^{2}-(\frac{5}{\sqrt{2}})^{2}-3=0$

$(2\sqrt{2}*x)^{2}-2*2\sqrt{2}*(\frac{5}{\sqrt{2}})+(\frac{5}{\sqrt{2}})^{2}-\frac{25}{2}-3=0$

$(2\sqrt{2}*x-\frac{5}{\sqrt{2}})^{2}-\frac{25+6}{2}}=0$

$(2\sqrt{2}x-\frac{5}{\sqrt{2}})^{2}=\frac{31}{2}}$