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2 years ago

Determine the nature of the solutions to the equation: x2 - 3x = 6

Mathematics

2 answers:

olga55 [171]2 years ago

8 0

Answer:

One Solution

Step-by-step explanation: when x is equal to one number it is defined as one solution. I hoped I answered correctly. please comment if I haven't

uysha [10]2 years ago

4 0

Answer: The solution is two distinct real zeros.

Step-by-step explanation:

Subtract 6 from both sides:

$x^2-3x-6=6-6\\x^2-3x-6=0$

Calculate the discriminant:

$D=b^2-4ac$

Where:

a=1

b=-3

c=-6

Then:

$D=(-3)^2-4(1)(-6)=33$

By definition, when $D>0$ the solution is two distinct real zeros.

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Read 2 more answers

Pleases someone explain

Nikolay [14]

When you have a negative exponent, you move the base with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

$\frac{1}{y^{-2}} =\frac{y^2}{1}$ or $y^2$

$2x^{-3}$ or $\frac{2^1x^{-3}}{1} =\frac{2}{x^3}$

$\frac{1}{x^{-5}} =x^5$

When a base with an exponent is divided by a base with an exponent, you subtract the exponents together. (But you can only combine the exponents when the bases are the same)

For example:

$\frac{x^2}{y}$ (can't combine because they have different bases of y and x)

$\frac{x^5}{x^3} =x^{(5-3)}=x^2$

$\frac{2^2}{2} =2^{(2-1)}=2^1 = 2$

When you multiply an exponent directly to a base with an exponent, you multiply the exponents together.

For example:

$(x^2)^3=x^{(2*3)}=x^6$

$(y^2)^{10}=y^{(2*10)}=y^{20}$

$\frac{2e^0}{(e^{-3})^2}$ First multiply the exponents together in the denominator

$\frac{2e^0}{e^{(-3*2)}}= \frac{2e^0}{e^{-6}}$ Now subtract the exponents together

$2e^{(0-(-6))}$ (two negative signs cancel each other out and become positive)

$2e^{(0+6)}$

$2e^6$

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2 years ago