Menu

A Shifted Square

Part of the Quadratics & Complex Numbers section of Coddy's Algebra 2 journey. Lesson 26 of 48.

When a squared bracket equals a negative, take the square root of both sides straight away. For (x−2)2=−9{\left(x-2\right)}^{2} = -9 that gives x−2=3ix-2 = 3i or x−2=−3ix-2 = -3i.

Each case is now a one-step equation: move the 2 across. So x=2+3ix = 2+3i or x=2−3ix = 2-3i, written with the real part first, as a+bia+bi.

Why not expand first? Expanding gives x2−4x+4=−9{x}^{2}-4x+4 = -9, a full quadratic, and you would have to complete the square all over again. The square is already complete, so the root goes straight to the answer. Check one: 2+3i−22+3i-2 is 3i, and (3i)2{\left(3i\right)}^{2} is -9.

challenge icon

Challenge

Solve (x+3)2=−16{\left(x+3\right)}^{2} = -16.

Try it yourself

(x+3)2=−16\left(x+3\right)^{2} = -16

Get xx alone

All lessons in Quadratics & Complex Numbers