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4x=square root 8+8x
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\[\huge{4x=\sqrt{8+8x}}\] \[\huge{16x^2=8+8x}\] \[\huge{16x^2-8x-8=0}\] \[\huge{a = 16 , b = -8,c =- 8 }\] can u solve it now ?
No..
with the help of quadratic equation : \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
with the help of this formula?
try to solve ..
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\[4x = \sqrt{8+8x}\] \[=> 16x^2 = 8+8x\] \[=> 2x^2 - x - 1 = 0\] \[=> 2x^2 - 2x + x - 1 = 0\] \[=> 2x(x - 1)+1(x-1) = 0\] \[=> (2x+1)(x-1) = 0\] \[=> x = \frac{-1}{2} or (1)\]
I honestly don't know how.
@mathslover I don't think there is any need for applying quadratic formula
I agree, you don't need to apply the quadratic formula here. You can if you want to, but unless you're a computer program there's methods that take less time.
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