How can I solve a disguised quadratic equation? 3/x+4 - 2/x = 5?
multiply every term by x, 3/x + 4 - 2/x = 5 x * [ 3/x + 4 - 2/x] = x * [5] 3 + 4x - 2 = 5x 1 + 4x = 5x 1 = x <==== ANSWER OR did you mean, 3/(x+4) - 2/x = 5? ... Then, get a common denomoninatior on the left hand side, 3x / (x * (x + 4)) - 2(x+4) / (x * (x + 4)) = [ 3x - 2x - 8 ] / [x² + 4x] = 5 [x - 8] / [x² + 4x] = 5 now multiply through by x² + 4x, x - 8 = 5[x² + 4x] x - 8 = 5x² + 20x simplify, 0 = 5x² + 19x + 8 now solve with the quadratic fromula to get: x = -19/10 ± √[201] / 10
Which of the following is not a disguised quadratic equation?
b) x^5 is not equal to (x^3)^2
How to solve this disguised quadratic equation?
The lowest common denominator is x^6 (not x^8): Multiply everything by x^6 to get rid of the denominators: 5x^4 = x^8 + 4 Now put everything on one side: x^8 - 5x^4 + 4 = 0 Next substitute u = x^4: u² - 5u + 4 = 0 Factor: (u - 1)(u - 4) = 0 u = 1 or u = 4 Substitute back for u: x^4 = 1 or x^4 = 4 Solutions for x^4 = 1 x = -1 x = 1 x = -i x = i Solutions for x^4 = 4 x = -√2 x = √2 x = (-√2)i x = (√2)I If they don't want the imaginary solutions, you have the following real solutions: x = {-1, 1, -√2, √2}
Which of the following is not a disguised quadratic equation?
b because you can't factor anything out of all three terms (except for 1).
Disguised quadratic equations?
for example take y=3x-1 the equation become y^2 + 6y - 7 = 0 factorrize: (y + 7)(y - 1) = 0, y = -7 and y = 1 replace back, 3x-1 = 7 ---> x = 8/3 3x-1 = 1, ---> x=2/3
Disguised Quadratic equations?
2p² + p - 10 = 0 (2p² - 4p) + (5p - 10) = 0 2p(p - 2) + 5(p - 2) = 0 (2p + 5)(p - 2) = 0 p = -5 / 2 or p = 2. If p = x + 1 / x, then 2p² + p - 10 = 0 implies 2(x + 1 / x)² + (x + 1 / x) - 10 = 0, 2(x² + 2 + 1 / x²) + (x + 1 / x) - 10 = 0, 2x² + 4 + 2 / x² + x + 1 / x - 10 = 0, 2x² + x - 6 + 1 / x + 2 / x² = 0. Multiplying through by x², 2x^4 + x³ - 6x² + x + 2 = 0. Therefore, 2x^4 + x³ - 6x² + x + 2 = 0 has solutions p = -5 / 2 or p = 2, where p = x + 1 / x. x + 1 / x = -5 / 2 implies 2x² + 2 = -5x 2x² + 5x + 2 = 0 (2x² + 4x) + (x + 2) = 0 2x(x + 2) + 1(x + 2) = 0 (2x + 1)(x + 2) = 0 x = -1 / 2 or x = -2. x + 1 / x = 2 implies x² + 1 = 2x x² - 2x + 1 = 0 (x - 1)² = 0 x - 1 = 0 x = 1. Therefore, x = -2 or x = -1 / 2 or x = 1.
Question about Disguised Quadratic equations (algebra)?
Here you are not doing anything to the terms of the equation, other than replace one of them by something that it equals. Start with x^4 + 5x^2 - 14 = 0, notice that x^4 EQUALS (x^2)^2. This is because (x^2)^2 = (x^2)(x^2) = x^4. You can replace anything by something that equals it. So replace the x^4 by (x^2)^2, and you then have the equivalent (and essentially the same) equation (x^2)^2+5(x^2)-14=0. Now if you give some other name, say u to x^2, the equation says u^2 + 5u - 14 = 0. You can solve that to get numeric values for u and then use the relationship u = x^2 to get x.
Use the quadratic formula to find the exact solution to: (x^4) - (5x^2) +3 = 0?
This is a quadratic in disguise. You can see it better with a temporary substitution for x². Let x² = p: x⁴ - 5x² + 3 = 0 (x²)² - 5x² + 3 = 0 p² - 5p + 3 = 0 By quadratic formula, p = (5±√13)/2 ➥ when p = (5+√13)/2, then it actually means x² = (5+√13)/2 → x = ±√[(5+√13)/2] ➥ when p = (5-√13)/2, then it actually means x² = (5-√13)/2→ x = ±√[(5-√13)/2] If you are allowed to use calculators and give decimal answers, then you have solutions x≈ ±2.0743, ±0.835
Solve the equation y −8y^1/2 +11 = 0, giving your answers in the form p±q√5.?
u ² - 8u + 11 = 0 u = [ - b ± √ ( b ² - 4 a c ) ] / 2 a u = [ 8 ± √ ( 64 - 44 ) ] / 2 u = [ 8 ± √ ( 20 ) ] / 2 u = [ 8 ± 2√5 ] / 2 u = [ 4 ± √5 ] y^(1/2) = [ 4 ± √5 ] y = [ 4 ± √5 ] [ 4 ± √5 ] y = 16 + 8√5 + 5 , y = 16 - 8√5 + 5 y = 21 + 8√5 , y = 21 - 8√5
Solve the equation: 8 square root x = 15 + x. (its a quadratic equation in disguise). i need help, thanks!?
x1 = 25 x2 = 9 3 very simple lines of calculation from basic school :)