Determine the number of solutions for each quadratic.

Mathematics

1 answer:

DiKsa [7]3 years ago

4 0

Answer:

1. Technically 2, but might be 0 in your teacher's opinion.

2. 1

Step-by-step explanation:

Solving problem one.

So I don't know if you have learned about imaginary numbers, but if you have, then you would end up with two answers if you plugged in the quadratic formula.

If you haven't learned about imaginary numbers, then I would say your best option would be to write 'No real solution' since there are technically 2 solutions.

Solving problem two.

Turns out this quadratic has a special property and it's actually a square of one equation. You can find out by just factoring the equation.

It's (3x-2)^2. Since it's squared, that means that only 2/3 would work as x in this equation.

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3 years ago

The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi

slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}$

Now for the initial condition the value of C is calculated as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16$

So the solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}$

Simplifying the equation as

$y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}$