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

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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 :)

Answer 4

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!

Answer 5

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 >>>>>>>>> ax^2+bx+c =0 as required

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

note that b^2 < 4ac, therefore the roots of the equation are complex

in general,

if b^2 > 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 < 4ac, the roots are complex numbers

i hope that this helps

Answer 6

5 + x = sqrt(3 + 4x)

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

x^2 + 6x + 22 = 0

Answer 7

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

Answer 8

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 ==>
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

Answer 9

each of here triples [a,b,c] is a answer: [-20, 60, 80 one], [-20, a hundred, -seventy one], [-18, ninety, -72], [-18, ninety, -8], [-sixteen, 80, -60], [-sixteen, 80, -15], [-15, 60, 4], [-14, 70, -20], [-12, 60, -40 seven], [-12, 60, -23], [-10, 50, -24], [-8, 40, -32], [-8, 40, -23], [-6, 30, -20], [-4, 20, -15], [-2, 10, -8], [2, -10, 12], [6, -30, 40], [8, -40, 40 8], [8, -40, fifty seven], [10, -50, seventy six], [12, -60, seventy 3], [12, -60, ninety seven] and of direction there are various extra. Edit: that's a curious limitless kin of ideas. for each integer n the polynomial f(x) = (a million/9)(a million-32*sixteen^n+40*4^n) x^2 + (a million/9)((a hundred and sixty*sixteen^n-152*4^n-8)) x + (sixteen-128*sixteen^n+112*4^n) has integer coefficients and f(a million) = a million^2, f(2) = ((8*4^n-2)/3)^2, f(3) = ((8*4^n+a million)/3)^2, f(4) = (4*2^n)^2 and f(5) = (a million/9)(a million-128*sixteen^n+352*4^n) that's unfavorable for n>0 and so not a sq.. e.g. n=5 supplies f(x) = -3723719*x^2+18624056*x-14900336 and f(a million)=a million^2, f(2)=2730^2, f(3)=2731^2, f(4)=128^2, and f(5)= -14873031

Answer 10

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