Answer:
Option B) [tex]6^2[/tex] is correct.
That is in the given equation [tex]x-12\sqrt{x}+36=0[/tex] the value of x is [tex]6^2[/tex]
ie., [tex]x=6^2[/tex]
Step-by-step explanation:
The given equation is [tex]x-12\sqrt{x}+36=0[/tex]
To find the value of x in the given equation :
[tex]x-12\sqrt{x}+36=0[/tex]
Rewritting the above equation as below :
[tex]\sqrt{x}^2-2(\sqrt{x})(6)+6^2=0[/tex]
Which is in the form of [tex](a-b)^2=a^2-2ab+b^2[/tex]
Here [tex]a=\sqrt{x}[/tex] and b=6
Therefore it can be written as
[tex](\sqrt{x}-6)^2=0[/tex]
[tex]\sqrt{x}-6=0[/tex]
[tex]\sqrt{x}=6[/tex]
Squaring on both sides we get
[tex](\sqrt{x})^2=6^2[/tex]
[tex]x=6^2[/tex]
Therefore in the given equation [tex]x-12\sqrt{x}+36=0[/tex] the value of x is [tex]6^2[/tex]
Therefore Option B) [tex]6^2[/tex] is correct.
