How do I solve this?
25x^4=74x^2-49
Thanks!
25x^4=74x^2-49
Thanks!
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How do I solve this?
25x^4=74x^2-49
this equation is in 'quadratic' form...
25x^4 - 74x^2 + 49 = 0
(25x^2 - 49)(x^2 - 1) = 0
check
both of the binomial factors are the differences of two squares, so...
(5x + 7)(5x - 7)(x + 1)(x - 1) = 0
x = - 7/5 or x = 7/5 or x = - 1 or x = 1
check
@Σ
25x^4=74x^2-49
this equation is in 'quadratic' form...
25x^4 - 74x^2 + 49 = 0
(25x^2 - 49)(x^2 - 1) = 0
check
both of the binomial factors are the differences of two squares, so...
(5x + 7)(5x - 7)(x + 1)(x - 1) = 0
x = - 7/5 or x = 7/5 or x = - 1 or x = 1
check
@Σ
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Use y=x^2.
The equation tranforms to 25y^2 = 74y - 49.
Use the quadratic formula, and solve for y. Then remember y = x^2, so take the positive and nehgative square root.
The equation tranforms to 25y^2 = 74y - 49.
Use the quadratic formula, and solve for y. Then remember y = x^2, so take the positive and nehgative square root.
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Well. You have to perform a lot of steps. It's not worth it. Just drop out.