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Worked solution

Explain why the new quadratic roots are not real

In secondary school, −1-1 and 11 are the roots of x2−1=0x^2-1=0. What are the roots of x2+1=0x^2+1=0? Prove that this equation has no real solution, then express its roots using i2=−1i^2=-1.

CJC · Lecture · Opening activity

Prove there is no real root

Answer

No real x satisfies x2+1=0x^2+1=0.

For every real x, x2≥0x^2\ge0, so x2+1≥1>0x^2+1\ge1>0. It therefore cannot equal zero. This supplies the proof requested in the opening source question.

How did this part go?

Use the imaginary unit to give both roots

Answer

x=ix=i or x=−ix=-i.

Rewrite the equation as x2=−1x^2=-1. With i2=−1i^2=-1, both i and −i square to −1. Equivalently, x2+1=(x−i)(x+i)x^2+1=(x-i)(x+i), so these are exactly its two roots. The source writes ±−1\pm\sqrt{-1} before introducing i; choosing the principal square root gives −1=i\sqrt{-1}=i, while the quadratic still has both signs.

How did this part go?