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Find solution of quadratic equation: 4x^2+16=52
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\(\Large \color{MidnightBlue}{\Rightarrow 4x^2 = 36 }\) \(\Large \color{MidnightBlue}{\Rightarrow 4x^2 - 36 = 0 }\) \(\Large \color{MidnightBlue}{\Rightarrow 4(x^2 - 9)=0 }\) \(\Large \color{MidnightBlue}{\Rightarrow x^2 - 9 = 0 }\) \(\Large \color{MidnightBlue}{\Rightarrow x^2 = 9 }\) \(\Large \color{MidnightBlue}{\Rightarrow x = 3 }\)
@parthkohli \[x=\pm 3 \]
Oh yes..
OR \[\large 4x^2 = 36\] \[\large \frac{4x^2}{4} = \frac{36}{4}\] \[\large x^2 = 9\] \[\large \sqrt{x^2} = \sqrt{9}\] \[\large x = \pm 3\] whichever you find easier
or \[4x^2-36=0\]\[4(x^2-9)=0\]\[4(x-3)(x+3)=0\]\[x=3,-3\]
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Thanks everyone :)
:D
OR . . . . . nahh i got nothing
lol ewgee
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