# Which equation has the solution x = 1 ± √5?

x^{2} + 2x + 4 = 0

x^{2} - 2x + 4 = 0

x^{2} + 2x - 4 = 0

x^{2} - 2x - 4 = 0

**Solution: **

**We will use the formula x = (−b ± √(b ^{2}− 4ac)) / 2a to find the roots of a quadratic equation ax^{2}+ bx + c = 0 and check all the equations.**

⇒ x^{2 }+ 2x + 4 = 0

x = (- 2 ± √(4 - 16)) / 2

= (-2 ± i√12) / 2

**This equation does not have the solution x = 1 ± √5.**

⇒ x^{2} - 2x + 4 = 0

x = (2 ± √(4 - 16)) / 2

=(2 ± i√12) / 2

**This equation does not have the solution x = 1 ± √5.**

⇒ x^{2} + 2x - 4 =0

x = (-2 ± √(4 + 16)) / 2

= (-2 ± √20) / 2

=(-2 ± √20) / 2

= (- 1 ± √5 )

**This equation does not have the solution x = 1 ± √5.**

⇒ x^{2} - 2x - 4 =0

x = (2 ± √(4 + 16)) / 2

= (2 ± √20) / 2

=(1 ± √5)

**This equation has the solution x = 1 ± √5.**

## Which equation has the solution x = 1 ± √5? x^{2} + 2x + 4 = 0, x^{2} - 2x + 4 = 0, x^{2} + 2x - 4 = 0, x^{2} - 2x - 4 = 0

**Summary:**

The equation which has the solution x = 1 ± √5 is x^{2} - 2x - 4 = 0.

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