# Roots of Quadratic Equation?

If r and s are the roots of 2x²+4x+3=0, what is the value of (1/r²)+(1/s²)?

### 4 Answers

- 12 months agoFavorite Answer
r = (b + sqrt(b^2 - 4ac)) / (2a)

s = (b - sqrt(b^2 - 4ac)) / (2a)

1/r^2 + 1/s^2 =>

(s^2 + r^2) / (r * s)^2

s^2 + r^2 =>

(1/(2a))^2 * ((b - sqrt(b^2 - 4ac))^2 + (b + sqrt(b^2 - 4ac))^2) =>

(1/(4a^2)) * (b^2 - 2b * sqrt(b^2 - 4ac) + b^2 - 4ac + b^2 + 2b * sqrt(b^2 - 4ac) + b^2 - 4ac)) =>

(1/(4a^2)) * (b^2 + b^2 + b^2 + b^2 - 4ac - 4ac + 2b * sqrt(b^2 - 4ac) - 2b * sqrt(b^2 - 4ac)) =>

(1/(4a^2)) * (4b^2 - 8ac) =>

(4b^2 - 8ac) / (4a^2) =>

(b^2 - 2ac) / a^2

r * s =>

(1/(2a)) * (1/(2a)) * (b - sqrt(b^2 - 4ac)) * (b + sqrt(b^2 - 4ac)) =>

(1/(4a^2)) * (b^2 - (b^2 - 4ac)) =>

(1/(4a^2)) * (b^2 - b^2 + 4ac) =>

4ac / (4a^2) =>

c/a

(r * s)^2 = c^2 / a^2

(r^2 + s^2) / (rs)^2 =>

((b^2 - 2ac) / a^2) / (c^2 / a^2) =>

(b^2 - 2ac) / c^2

Now you have a general formula for any question structured like this

a = 2 , b = 4 , c = 3

(4^2 - 2 * 2 * 3) / 3^2 =>

(16 - 12) / 9 =>

4/9

- Anonymous12 months ago
The right side of your equation is prime, that is, it does not factor over the set of rationals. Since the discriminant is D = 4^2 - 4(2)(3) = -8, both r and s are complex conjugates. As such,

(1/r²)+(1/s²) = 1/r² + 1/[conj(r)]²

= [r² + (conj(r))²] / |r|⁴

= 2|r|²/|r|⁴

= 2/|r|².

Solving the equation, we may take r = -1 + I/sqrt(2); Thus, |r|² = 5/4. Hence the quantity we want is 2/(5/4) = 8/5.

- atsuoLv 612 months ago
2x^2 + 4x + 3 = 0

x^2 + 2x + 3/2 = 0

r and s are the roots of this equation , so

(x - r)(x - s) = 0

x^2 - (r + s)x + rs = 0

So

r + s = -2 and rs = 3/2

Therefore

1/r^2 + 1/s^2

= (s^2 + r^2) / (r^2s^2)

= [(r + s)^2 - 2rs] / (rs)^2

= [(-2)^2 - 2*(3/2)] / (3/2)^2

= 1 / (9/4)

= 4/9

- Steve ALv 712 months ago
x = (-4 +/- √(16 -4*3*2))/2*2

x = (-4 +/-√(-8))/4

x = -1 + i√(2)/2 ; -1 - i√(2)/2

Let r be the + version and s be the - version.

r^2 = 1/2 +i√2

s^2 = 1/2 - i√2

1/r^2 + 1/s^2 = 4/9