Convert "5 + x - sqrt(3+4x) = 0" into standard quadratic form, ie "ax^2 + bx + c = 0"?

Can you show every step in your working please.





I know the answer (it's in the back of the book) but I can't quite get it!

Convert "5 + x - sqrt(3+4x) = 0" into standard quadratic form, ie "ax^2 + bx + c = 0"?
Isolate the sqrt term, so you have 5 + x = sqrt(3 + 4x). Then square both sides, giving you x^2 + 10x + 25 = 3 + 4x. Combine terms to get x^2 + 6x + 22 = 0.
Reply:OK, here we go step by step:





(a) 5 + x - sqrt (3 + 4x) = 0





(b) 5 + x = sqrt (3 + 4)





(c) (5 + x) ^2 = 3 + 4x Note: ^2 means squared





(d) (5 + x)(5 + x) = 3 + 4x





(d) 5(5 + x) + x(5 + x) = 3 + 4x





(e) 25 + 5x + 5x + x^2 = 3 + 4x





(f) x^2 + 10x + 25 = 3 + 4x





(g) x^2 +10x - 4x + 25 - 3 = 0





(h) x^2 + 6x + 22 = 0





Which is now in the form of the general quadratic





where a = 1, b = 6 and c = 22
Reply:Do your own homework
Reply:thats simple. i wont bore you with the answer, mind.





instead try to guess how many fingers im holding up.








Not as easy as it sounds is it? back to the books 4 u
Reply:5 + x - sqrt(3+4x) = 0------square this eqn


25+10x+ x^2 - (3+4x)=0


so, x^2 + 6X + 22=0


thus, a=1,b=6,c=22
Reply:5+x -sqrt(3+4x) = 0





move sqrt(3+4x) to rhs of equation





(5+x) = sqrt (3+4x)





square each side





(5+x)^2 =(3+4x)





x^2 +10x +25 = 3+ 4x





move 3+4x back to lhs of equation





x^2+6x+22 = 0 %26gt;%26gt;%26gt;%26gt;%26gt;%26gt;%26gt;%26gt;%26gt; ax^2+bx+c =0 as required





giving a=1,b=6 and c=22





note that b^2 %26lt; 4ac, therefore the roots of the equation are complex





in general,





if b^2 %26gt; 4ac, the roots of the equation are real and unequal





if b^2 = 4ac, the roots of the equation are real and equal





if b^2 %26lt; 4ac, the roots are complex numbers





i hope that this helps
Reply:5 + x - sqrt(3 + 4x) = 0


-sqrt(3 + 4x) = - 5 - x


sqrt(3 + 4x) = 5 + x





Squaring on both sides and simplifying


3 + 4x = 5^2 + 2 * 5 * x + x^2


3 + 4x = 25 + 10x + x^2


x^2 + 6x + 22 = 0





which is in the standard quadratic equation form ax^2 + bx + c = 0





with a = 1, b = 6 and c = 22





Hope it helps :)
Reply:could try squaring everything





so





25 + x^2 - 3 +4x = 0


so x^2 + 4x + 22 = 0





i am probably wrong, what is the answer??
Reply:Hi there!





So, the first thing you have to do is to push sqrt part to the right hand side


5+x = sqrt(3+4x)





since sqrt(x)= x raised to (1/2)


and it is not possible (as far as i know) to expand sqrts,





(just to show ==%26gt;


sqrt(4) = 2


4 ^ (1/2) = 2


4 = 2 ^ 2 ... ^ means power)





so,


5+x = sqrt(3+4x) becomes


(5+x) ^2 = 3 + 4x


x^2 + 10x + 25 = 3 + 4x


x^2 +6x + 22





cheers
Reply:5 + x = sqrt(3 + 4x)





25 + 10x + x^2 = 3 + 4x





x^2 + 6x + 22 = 0
Reply:5 + x - sqrt(3+4x) = 0


5 + x = sqrt(3+4x)


(5 + x)^2 = 3 + 4x


25 + 10x + x^2 = 3 + 4x


25 + 10x + x^2 - 3 - 4x = 0


x^2 + 6x + 22 = 0
Reply:You have to (well you don't but it's the easiest way) put the square root on the right side of the equation:


"5 + x = sqrt (3+4x)"


then square it...remember (a + b)^2 = a^2 + 2ab + b^2 so


"25 + 10x + x^2 = 3 + 4x" and


"x^2 + 6x + 22 = 0"





Should be right!
Reply:5+x = sqrt(3+4x) multiply both sides by (3+4x)





(5+x)(3+4x)=(3+4x)


15 + 3x + 20x + 4xsqrd = 3 + 4x





4xsqrd + 19x + 12 = 0
Reply:5 + x - sqrt(3 + 4x) = 0





x - sqrt(3+4x) = -5





squaring both sides, we get





x^2 - 2xsqrt(3+4x) + (3 + 4x) = 25





x^2 - 2xsqrt(3 + 4x) + 3 + 4x = 25





x^2 - 2x[sqrt(3 + 4x) - 2x] -22 = 0





a = 1, b = 2[sqrt(3 + 4x) - 2x, c = -22
Reply:this sounds suspiciously like homework... what would your teacher say?!