更新1:
With steps please
Is someone able to put this equation in quadratic form? 4x+5=8√(1-x^2 )?
回答 (3)
2016-05-25 6:52 am
✔ 最佳答案
First, square both sides:∴ (4x + 5)^2 = (8√(1-x^2))^2
∴ 16x^2 + 40x + 25 = 64 (1-x^2)
∴ 16x^2 +40x + 25 = 64 - 64x^2
Now rearrange to get the required quadratic equation;
∴ 80x^2 + 40x - 39 = 0
Hope this helps !!!!!!!!
2016-05-25 6:56 am
4x + 5 = 8√(1 - x² )
[4x + 5]² = [8√(1 - x² )]²
16x² + 40x + 25 = 64(1 - x²)
16x² + 40x + 25 = 64 - 64x²
80x² + 40x - 39 = 0
x = {-40 ± √[40² - 4(80)(-39)]} / [2(80)]
x = (-5 + 2√55)/20 or x = (-5 - 2√55)/20
[4x + 5]² = [8√(1 - x² )]²
16x² + 40x + 25 = 64(1 - x²)
16x² + 40x + 25 = 64 - 64x²
80x² + 40x - 39 = 0
x = {-40 ± √[40² - 4(80)(-39)]} / [2(80)]
x = (-5 + 2√55)/20 or x = (-5 - 2√55)/20
2016-05-25 7:26 am
First square both sides getting 16x^2 + 40x + 25 = 64 - 64x^2. Then 80x^2 + 40x - 39 = 0.
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