Answer:
[tex]\displaystyle \large{x=\pm \sqrt{y}}[/tex]
Step-by-step explanation:
We are given the equation:
[tex]\displaystyle \large{y=x^2}[/tex]
Make ‘x’ the subject of the formula by square root both sides:
[tex]\displaystyle \large{\sqrt{y}=\sqrt{x^2}}[/tex]
Cancel the square root and square in RHS and write plus-minus in LHS:
[tex]\displaystyle \large{\pm \sqrt{y}= x}\\\displaystyle \large{x=\pm \sqrt{y}}[/tex]
Therefore, the solution is [tex]\displaystyle \large{x=\pm \sqrt{y}}[/tex]
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Summary
To make a variable as the subject of equation means to isolate that variable. For an example, if you want to make y as the subject, you’ll have to isolate y-term.
You may also be curious why do you have to add plus-minus on another side when cancelling [tex]\displaystyle \large{\sqrt{x^2}}[/tex] because [tex]\displaystyle \large{\sqrt{x^2}=|x|}[/tex] and [tex]\displaystyle \large{|x|}[/tex] for their positive and their counterparts (opposite/negative) will always give same y-value.
From an example:
[tex]\displaystyle \large{\sqrt{x^2}=4}[/tex] can be rewritten as [tex]\displaystyle \large{|x|=4}[/tex], cancel absolute sign/square root then write plus-minus which the solution is [tex]\displaystyle \large{x=\pm 4}[/tex] and if you substitute x = 4 or -4, the equation will be true for both values.
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Others
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