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2 years ago

Consider the quadratic equation x2 = 4x - 5. How many solutions does the equation have?

Mathematics

1 answer:

JulijaS [17]2 years ago

3 0

Answer:

no real solutions
2 complex solutions

Step-by-step explanation:

The equation can be rearranged to vertex form:

x^2 -4x = -5 . . . . . . . . . subtract 4x

x^2 -4x +4 = -5 +4 . . . . add 4

(x -2)^2 = -1 . . . . . . . . . show the left side as a square

x -2 = ±√-1 = ±i . . . . . . take the square root; the right side is imaginary

x = 2 ± i . . . . . . . . . . . . . add 2. These are the complex solutions.

_____

Comment on the question

Every 2nd degree polynomial equation has two solutions. They may be real, complex, or (real and) identical. That is, there may be 0, 1, or 2 real solutions. This equation has 0 real solutions, because they are both complex.

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2. Express the given statement as an algebraic expression. Sum ofa and 5 divided by 8

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Answer:

An algebraic expression for Sum of a and 5 divided by 8 is $\mathbf{\frac{a+5}{8}}$

Step-by-step explanation:

We need to express the given statement as an algebraic expression.

Sum of a and 5 divided by 8

Basically, we are given the expression in English sentence and we have to write algebraic expression.

It can be done in two steps:

Sum of a and 5 = We know that sum is denoted by + sign. So, we get (a+5)

Sum of a and 5 divided by 8 = $\frac{a+5}{8}$

So, An algebraic expression for Sum of a and 5 divided by 8 is $\mathbf{\frac{a+5}{8}}$

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago

Simplify the following expression and show your work in details please .

wel

Answer:

9-2b

Step-by-step explanation:

Set up equation:

$9 - 8b + 6b$