Find x: (4x^2-25)^2-9(2x-5)^2=0

Mathematics

1 answer:

Feliz [49]2 years ago

6 0

Answer:

$(4 {x}^{2} - 25)^{2} - 9(2x - 5)^{2} = 0 \\ \\ {(4 {x}^{2} - 25)} (4 {x}^{2} - 25) - 9(2x - 5)(2x - 5) = 0 \\ \\( 16 {x}^{4} - 200 {x}^{2} + 625) - 9(4 {x}^{2} - 20x + 25) = 0 \\ \\ (16 {x}^{4} - 200 {x}^{2} + 625) + ( - 36 {x}^{2} + 180x - 225) = 0 \\ \\ 16 {x}^{4} - 236 {x}^{2} + 180x + 400 = 0 \\ \\ 4(x + 1)(x + 4)(2x - 5)^{2} = 0 \\ \\ x + 1 = 0 \\ x = - 1 \\ \\ x + 4 = 0 \\ x = - 4 \\ \\ {(2x - 5)}^{2} = 0 \\ 2x - 5 = 0 \\ 2x = 5 \\ x = \frac{5}{2}$

I hope I helped you^_^

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