Which equation has the solutions x=1+- square root of 5 A. x2 + 2x + 4 = 0 B. x2 – 2x + 4 = 0 C. x2 + 2x – 4 = 0 D. x2 – 2x – 4 = 0

Question

asked Feb 21, 2018 22.6k views

2 Answers

Rhinosforhire · Answer 1 · 2018-02-26T06:32:24+0000

Answer:

$x = 1\pm √(5)$

Explanation:

Equation:
ax^2+bx+c=0 --1

Quadratic formula :
$x = (-b\pm √(b^2-4ac))/(2a)$

Now Solve the given option using this formula

A)
x^2 + 2x + 4 = 0

On comparing with 1

a = 1 , b = 2 , c= 4

So,
$x = (-2\pm √(2^2-4(1)(4)))/(2(1))$

$x = (-2\pm √(-12))/(2)$

B)
x^2 - 2x + 4 = 0

On comparing with 1

a = 1 , b = -2 , c= 4

So,
$x = (2\pm √((-2)^2-4(1)(4)))/(2(1))$

$x = (2\pm √(-12))/(2)$

C)
x^2 + 2x - 4 = 0

On comparing with 1

a = 1 , b = 2 , c=- 4

So,
$x = (-2\pm √(2^2-4(1)(-4)))/(2(1))$

$x = (-2\pm √(20))/(2)$

$x = (-2\pm 2√(5))/(2)$

$x =-1\pm √(5)$

D)
x^2 - 2x - 4 = 0

On comparing with 1

a = 1 , b = -2 , c= -4

So,
$x = (2\pm √((-2)^2-4(1)(-4)))/(2(1))$

$x = (2\pm √(20))/(2)$

$x = (2\pm 2√(5))/(2)$

$x = 1\pm √(5)$

Hence Option D is correct.

x^2 - 2x - 4 = 0 has a solution
$x = 1\pm √(5)$