Solve the following system of equations: $ \begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases} $

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

$$ x=\sqrt{61}-1 $$ $$ y=\sqrt{61}+1 $$ or $$ x=\sqrt{61}+1 $$ $$ y=\sqrt{61}-1 $$

$$ x=0 $$ $$ y=\sqrt{61} $$ or $$ x=\sqrt{61} $$ $$ y=0 $$

$$ x=\sqrt{61}-2.5 $$ $$ y=\sqrt{61}+2.5 $$ or $$ x=\sqrt{61}+2.5 $$ $$ y=\sqrt{61}-2.5 $$

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

Solve the following system of equations: $ \begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases} $

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

$$ x=\sqrt{61}+5 $$ $$ y=\sqrt{61}-5 $$ or $$ x=\sqrt{61}-5 $$ $$ y=\sqrt{61}+5 $$

$$ x=0 $$ $$ y=\sqrt{61} $$ or $$ x=\sqrt{61} $$ $$ y=0 $$

$$ x=\sqrt{3}+1 $$ $$ y=\sqrt{3}-1 $$ or $$ x=\sqrt{3}-1 $$ $$ y=\sqrt{3}+1 $$

Solution of a system of equations - one of them is linear and the other quadratic

Solving a system of equations when one of them is linear and the other is quadratic

When we have a system of equations where one of the equations is linear and the other quadratic
we will use the substitution method:

We will isolate one variable from an equation, place in the second equation the value of the expression of the variable we have isolated, and in this way, we will obtain an equation with one variable. We will solve for $X$ or $Y$ and then place it in one of the original equations to find the complete point. The point we discover will be the point of intersection of the line with the parabola, and it will also be the solution of the system of equations.

Detailed explanation

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Test yourself on linear-quadraric systems of equations!

Choose the formula that represents line 1 in the graph below:

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Let's look at an example

Given the following system of equations: $y=x^2+2x+7$ , $2x+y=4$

Solution:
Let's isolate one of the variables from one of the equations:
$2x+y=4$
$y=-2x+4$

Now, let's substitute the value we isolated into the second equation. We will obtain:

$-2x+4=x^2+2x+7$

Let's combine like terms and transpose members. We will obtain:
$-X^2-4x-3=0$
We will solve it with the help of the quadratic formula and will obtain:
$X=-1 ,-3$
Now we will gradually place an $X$ we found into one of the equations to find the complete points: Let's start with $X=-1$

$2\times(-1)+y=4$
$-2+y=4$
$y=6$
Let's write this solution $(-1,6)$
let's move to the second $X$ we found $X=-3$ and we will obtain:
$2*(-3)+y=4$
$-6+y=4$
$y=10$ let's write this solution $(-3,10)$

These solutions are the points of intersection of the parabola with the axis.

If you are interested in this article, you might also be interested in the following articles:

The functions y=x²

Family of parabolas y=x²+c: Vertical shift

Family of parabolas y=(x-p)²

Family of parabolas y=(x-p)²+k (combination of horizontal and vertical shift)

Vertex form of the quadratic function

Factored form of the quadratic function

Completing the square in a quadratic equation

Standard form of the quadratic function

System of quadratic equations - Algebraic and graphical solution

In the blog of Tutorela you will find a variety of articles about mathematics.

Examples and exercises with solutions of linear and quadratic equation systems

Exercise #1

The following functions are graphed below:

$f(x)=x^2-6x+8$

$g(x)=4x-17$

For which values of x is
$f(x)<0$ true?

Video Solution

Step-by-Step Solution

To solve the inequality $f(x) = x^2 - 6x + 8 < 0$ , we follow these steps:

Step 1: Find the roots of the equation $f(x) = 0$ using the factoring method.
Step 2: Determine the intervals formed by these roots.
Step 3: Test points from each interval to determine where $f(x) < 0$ .

Step 1: Factor the quadratic equation $x^2 - 6x + 8 = 0$ .
Factoring gives: $(x - 2)(x - 4) = 0$ .

Thus, the roots are $x = 2$ and $x = 4$ .

Step 2: The roots divide the number line into three intervals: $x < 2$ , $2 < x < 4$ , and $x > 4$ .

Step 3: Choose a test point from each interval and plug it into $f(x)$ :

For $x < 2$ , choose $x = 1$ : $f(1) = 1^2 - 6(1) + 8 = 1 - 6 + 8 = 3$ , which is positive, so $f(x) \geq 0$ .
For $2 < x < 4$ , choose $x = 3$ : $f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1$ , which is negative, so $f(x) < 0$ .
For $x > 4$ , choose $x = 5$ : $f(5) = 5^2 - 6(5) + 8 = 25 - 30 + 8 = 3$ , which is positive, so $f(x) \geq 0$ .

Therefore, the interval where $f(x) < 0$ is $2 < x < 4$ .

The correct choice is:

$2 < x < 4$

Answer

$2 < x < 4$

Exercise #2

Look at the graph below of the following functions:

$f(x)=x^2-6x+8$

$g(x)=-x+4$

For which values of x is
$g(x)>0$ true?

Video Solution

Step-by-Step Solution

To solve the problem of finding for which values of $x$ , the function $g(x) = -x + 4$ is greater than zero, we begin as follows:

Step 1: Set up the inequality $g(x) > 0$ . This translates to $-x + 4 > 0$ .
Step 2: Solve the inequality:
- Subtract 4 from both sides: $-x > -4$ .
- Multiply both sides by $-1$ , reversing the inequality: $x < 4$ .

Therefore, the solution to the problem is that $g(x) > 0$ when $x < 4$ .

The corresponding choice that reflects this solution is choice 4: $x < 4$ .

Answer

$x<4$

Exercise #3

The following function is graphed below:

$f(x)=x^2-6x+8$

$g(x)=-x+4$

For which values of x is
$f(x) > g(x)$ true?

Video Solution

Step-by-Step Solution

To determine for which values of $x$ the condition $f(x) > g(x)$ holds, follow these steps:

Step 1: Set up the inequality $x^2 - 6x + 8 > -x + 4$ .
Step 2: Rearrange all terms to one side: $x^2 - 6x + 8 + x - 4 > 0$ simplifies to $x^2 - 5x + 4 > 0$ .
Step 3: Solve the related quadratic equation $x^2 - 5x + 4 = 0$ . This factors as $(x - 1)(x - 4) = 0$ , giving roots $x = 1$ and $x = 4$ .
Step 4: Determine intervals defined by the roots: $(-\infty, 1)$ , $(1, 4)$ , and $(4, \infty)$ .
Step 5: Test points from each interval in the inequality $x^2 - 5x + 4 > 0$ :
- For $x = 0$ (from interval $(-\infty, 1)$ ), $x^2 - 5x + 4 = 4$ which is greater than 0.
- For $x = 2$ (from interval $(1, 4)$ ), $x^2 - 5x + 4 = -2$ which is less than 0.
- For $x = 5$ (from interval $(4, \infty)$ ), $x^2 - 5x + 4 = 9$ which is greater than 0.
Step 6: From the test points, determine $f(x) > g(x)$ in intervals $(-\infty, 1)$ and $(4, \infty)$ .

Thus, $f(x) > g(x)$ for $x < 1$ or $x > 4$ .

Therefore, the solution to the problem is $x < 1, 4 < x$ , corresponding to choice 2.

Answer

$x < 1,4 < x$

Exercise #4

Choose the formula that describes graph 1:

Video Solution

Step-by-Step Solution

To solve this problem, we need to determine whether the provided graph corresponds to a quadratic or linear function.

First, we observe the shape of the graph.
The graph shows a downward curve, indicating it is a parabola.
The general form of a quadratic equation $(y = ax^2 + bx + c)$ implies graph types.
Let's verify if one of the given quadratic choices represents this graph.

Since the graph is a parabola opening upwards, we'll evaluate the given quadratic equation $y = x^2 - 6x + 8$ . Analyzing it and comparing leads to:

The vertex form can be rewritten or identified mathematically or visually from the given expression.
This quadratic formula aligns with prominent features of the parabola: its vertex, orientation, and intercepts, matching the graph.

Thus, as the parabola aligns perfectly with quadratic properties such as opening upwards, the formula that describes graph 1 correctly is:
$y = x^2 - 6x + 8$ .

Answer

$y=x^2-6x+8$

Exercise #5

The following functions are graphed below:

$f(x)=x^2-6x+8$

$g(x)=4x-17$

For which values of x is

$f(x)>0$ true?

Video Solution

Step-by-Step Solution

To solve the inequality $f(x) > 0$ , we first need to find the roots of the equation $f(x) = x^2 - 6x + 8$ .

1. Find the roots of the quadratic equation:
The quadratic is $x^2 - 6x + 8$ . This can be factored into:
$f(x) = (x - 2)(x - 4) = 0$ .

2. Calculate the roots:
Setting each factor equal to zero gives the roots $x = 2$ and $x = 4$ .

3. Determine the intervals defined by these roots:
The roots divide the x-axis into three intervals: $(-\infty, 2)$ , $(2, 4)$ , and $(4, \infty)$ .

4. Test points in each interval to decide positivity:
- For $x < 2$ , select $x = 1$ : $f(1) = 1^2 - 6(1) + 8 = 3 > 0$ . Thus, $f(x) > 0$ in $(-\infty, 2)$ .
- For $2 < x < 4$ , select $x = 3$ : $f(3) = 3^2 - 6(3) + 8 = -1 < 0$ . Thus, $f(x) < 0$ in $(2, 4)$ .
- For $x > 4$ , select $x = 5$ : $f(5) = 5^2 - 6(5) + 8 = 3 > 0$ . Thus, $f(x) > 0$ in $(4, \infty)$ .