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3 years ago

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Find the ordered pairs. {Y=(x+1)^2-5 {Y=2x PLEASE HELP!!!

1 answer:

vampirchik [111]3 years ago

6 0

Answer:

$\large \boxed{\sf \bf \ \ \text{The solutions are }(2,4) \text{ and } (-2,-4) \ \ }$

Step-by-step explanation:

Hello, this is the same kind of question that you have already posted( brainly.com/question/17141820).

<em>I would suggest that you check my previous answer and try to apply the same logic here. When you try it if you have any question please let me know. It will be the more efficient way for you to learn how to solve it by yourself.</em>

<em>As I have already provided details for the previous question I will be more succinct here.</em>

Again we can write y = y and then we have to solve the following.

$(x+1)^2-5=2x\\\\x^2+2x+1^2-5-2x=0\\\\x^2-4=x^2-2^2=(x-2)(x+2)=0\\\\x=2 \ \ or \ \ x=-2$

The solutions are (2,4) and (-2,-4)

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Tyrone solved an algebraic equation correctly. He got the answer 5 = 3. This answer indicates that the solution to the equation

worty [1.4K]

Answer:

No solution.

Step-by-step explanation:

5 can't equal 3 my friend.

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Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

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$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

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

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

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Answer:

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do addition and you should get

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