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3 years ago

64 POINT QUESTION √(5-2x) - √(2-x) = 1

Mathematics

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

x = 2 and x = -2

Step-by-step explanation:

The easiest way is to guess. We know that you can't have a square root of a negative number. This constrains x between negative infinity to 2. Testing 2 and -2, we see that it satisfy the equation.

The more systematic approach is shown below

$\sqrt{5-2x}-\sqrt{2-x}=1 \\\sqrt{5-2x}-\sqrt{2-x}+\sqrt{2-x}=1+\sqrt{2-x} \\\sqrt{5-2x} = 1+\sqrt{2-x}$

$5-2x=(1+\sqrt{2-x})^2\\5-2x=1+2\sqrt{2-x}+2-x\\2-x=2\sqrt{2-x}\\1-0.5x=\sqrt{2-x}\\(1-0.5x)^2=2-x\\1-x+0.25x^2=2-x\\0.25x^2-1=0\\0.25x^2 = 1\\x^2 = 4\\x = +/-2$

x= -2, 2

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Read 2 more answers

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$