How to solve for "y" | Oxford math admission test

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[@marcfirst9341]

There is no square roots of negative numbers in the real numbers, you to work a little bit on that, you can't just square both sides, there are some conditions to do that...

[@renesperb]

I would write the equation as√y *(1+i) = 4 . Then √y = 4/(1+i) = 4 (1-i)/2 = 2(1-i) -> y= 4(1-i)^2 = 8 i.

[@MathBeast.channel-l9i]

Brilliant approach Boss 😊

[@Mortadelo_]

yup it's way easier like that

[@colinkwan8604]

With this equation, the answer is -8i. Both -8i and 8i should be included as the answer.

[@colinkwan8604]

√y*(1+i)=4, √y=4/(1+i), y=(4/(1+i))^2...at the end you will get y=8/i, but this again just means y=-8i

[@Ryon007]

Ele😊

[@RealQinnMalloryu4]

(2)+(2) =4(y ➖ 2y+2).

[@mathmachine4266]

x=√(y).
√(-y)=±xi
4=x(1±i)
4(1-±i)/2=x
Differentiating between ± and -± here doesn't really matter
x=2(1±i)
y=x²=±8i
y=8i: √(y)=2+2i, √(-y)=2-2i
y=-8i: √(y)=2-2i, √(-y)=2+2i
y=±8i

[@martinfenner3222]

if you really want to pass the Oxford math admission test, you should improve your solution in several ways:
(1) as mentioned by others, you missed the second solution y=-8i. Take into account, that every equation of the form x² = T always has two solutions +/- sqrt (T)
(2) IMHO in the verification calculation you cannot square both sides of the equation, because of the same reason. This is not an equivalence transformation. Otherwise you verify -1 = 1 just by squaring both sides of that equation to (-1)² = 1². Instead, You should calculate sqrt (+/-8i) directly, maybe by an approach like (a+ib)² = +/8i. Calculating a and b yields the desired squareroots.
(3) you can simplify your calculation by squaring sqrt (y) + sqrt (-y) = 4 right from the beginning. By careful evaluation of the resulting expressions you should obtain the same result +/-8i with much less steps.
I hope, this helps to pass the test. Good luck.

[@AytuğSönmezoğlu]

good vid.🙂

[@Steven-v6l]

correct answers 8i and -8i, because
8i ==> √(8i) + √(-8i)
-8i ==> √(-8i) + √(-(-8i)) = √(-8i) + √(8i) = √(8i) + √(-8i)
other than that ... your calculation was correct

[@NewsTopThe]

Thank you.

[@MathBeast.channel-l9i]

You are welcome 🤗

[@bucharipts9143]

You miss another one, solution.
√64*(-1) = 8i or -8i
So y = 8i or y = -8i

[@joaquimcosta8107]

8i e -8i