Find the real solutions of 2x^4+10x^2=48...?

2x^4 + 10x² - 48 = 0

=> 2(x^4 + 5x² - 24) = 0

=> 2(x² + 8)(x² - 3) = 0

=> (x² + 8)=0 which has no real solutions
or (x² - 3)=0

=>x²=3
=>x=±√3

Cheers.....

2x^4 + 10x² - 48 = 0
x^4 + 5x² - 24 = 0
(x² + 8)(x² - 3) = 0
x² = - 8 , x² = 3
x² = 8i² , x² = 3
x = ± i 2√2 , x = ± √3

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2x^4 + 10x^2 = 48
2x^4 + 10x^2 - 48 = 0
(2x^2 - 6)(x^2 + 8) = 0

2x^2 - 6 = 0
2x^2 = 6
x^2 = 3
x = ± √3

x^2 + 8 = 0
x^2 = -8
x = √-8 (imaginary number)

∴ x = ± √3

2x^4+10x^2=48
x^4+5x^2-24=0
(x^2-3)(x^2+8)=0
x^2-3=0, x^2+8=0
x^2=3, x^2=-8 these solutions will not be real
x= +,- sqrt(3)

2x^4+10x^2=48
(2x^2+16)(x^2-3)=0
x^2=3
x= positive or negative sqrt of 3





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Let y=x^2 then
2y^2+10y=48
=> 2y^2+10y-48=0
=> y^2+5y-24=0
=> (y+8)(y-3)=0
=> y=-8 or 3

But y =x^2 cannot be negative (otherwise you get complex solution for x)

So x^2 = 3
=> x = +/-sqrt(3)

2x^4 + 10x^2 - 48 = 0
2(x^4 + 5x^2 - 24) = 0
2(x^2 + 8)(x^2 - 3) = 0
since you want real solutions, you may discard x^2 = -8 since this would be imaginary.
Then x^2 = 3 so x = +/- sqrt 3