Question Solve $10x + y = -x^2 + 2$ for real solutions. Determine if there are 0, 1, or 2 real solutions.

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $10x + y = -x^2 + 2$ .
2. We need to determine the number of real solutions to this equation.

STEP 2

To find the number of real solutions, we can treat the equation as a quadratic equation in terms of $x$ by isolating $y$ .
$y = -x^2 - 10x + 2$

STEP 3

The equation $y = -x^2 - 10x + 2$ is a quadratic equation in the standard form $y = ax^2 + bx + c$ , where $a = -1$ , $b = -10$ , and $c = 2$ .

STEP 4

To determine the number of real solutions, we can use the discriminant of the quadratic equation, which is given by $\Delta = b^2 - 4ac$ .

STEP 5

Calculate the discriminant using the values of $a$ , $b$ , and $c$ .
$\Delta = (-10)^2 - 4(-1)(2)$

STEP 6

Simplify the discriminant.
$\Delta = 100 - (-8)$ $\Delta = 100 + 8$ $\Delta = 108$

STEP 7

Since the discriminant $\Delta > 0$ , the quadratic equation has two distinct real solutions.
The equation $10x + y = -x^2 + 2$ has two real solutions.

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Question Solve $10x + y = -x^2 + 2$ for real solutions. Determine if there are 0, 1, or 2 real solutions.

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $10x + y = -x^2 + 2$ .
2. We need to determine the number of real solutions to this equation.

STEP 2

To find the number of real solutions, we can treat the equation as a quadratic equation in terms of $x$ by isolating $y$ .
$y = -x^2 - 10x + 2$

STEP 3

The equation $y = -x^2 - 10x + 2$ is a quadratic equation in the standard form $y = ax^2 + bx + c$ , where $a = -1$ , $b = -10$ , and $c = 2$ .

STEP 4

To determine the number of real solutions, we can use the discriminant of the quadratic equation, which is given by $\Delta = b^2 - 4ac$ .

STEP 5

Calculate the discriminant using the values of $a$ , $b$ , and $c$ .
$\Delta = (-10)^2 - 4(-1)(2)$

STEP 6

Simplify the discriminant.
$\Delta = 100 - (-8)$ $\Delta = 100 + 8$ $\Delta = 108$

STEP 7

Since the discriminant $\Delta > 0$ , the quadratic equation has two distinct real solutions.
The equation $10x + y = -x^2 + 2$ has two real solutions.

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Question Solve 10x+y=−x2+210x + y = -x^2 + 210x+y=−x2+2 for real solutions. Determine if there are 0, 1, or 2 real solutions.

Studdy Solution

STEP 1

Assumptions1. We are given the equation 10x+y=−x2+210x + y = -x^2 + 210x+y=−x2+2.2. We need to determine the number of real solutions to this equation.

STEP 2

To find the number of real solutions, we can treat the equation as a quadratic equation in terms of xxx by isolating yyy.y=−x2−10x+2y = -x^2 - 10x + 2y=−x2−10x+2

STEP 3

The equation y=−x2−10x+2y = -x^2 - 10x + 2y=−x2−10x+2 is a quadratic equation in the standard form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c, where a=−1a = -1a=−1, b=−10b = -10b=−10, and c=2c = 2c=2.

STEP 4

To determine the number of real solutions, we can use the discriminant of the quadratic equation, which is given by Δ=b2−4ac\Delta = b^2 - 4acΔ=b2−4ac.

STEP 5

Calculate the discriminant using the values of aaa, bbb, and ccc.Δ=(−10)2−4(−1)(2)\Delta = (-10)^2 - 4(-1)(2)Δ=(−10)2−4(−1)(2)

STEP 6

Simplify the discriminant.Δ=100−(−8)\Delta = 100 - (-8)Δ=100−(−8) Δ=100+8\Delta = 100 + 8Δ=100+8 Δ=108\Delta = 108Δ=108

STEP 7

Since the discriminant Δ>0\Delta > 0Δ>0, the quadratic equation has two distinct real solutions.The equation 10x+y=−x2+210x + y = -x^2 + 210x+y=−x2+2 has two real solutions.

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Question Solve $10x + y = -x^2 + 2$ for real solutions. Determine if there are 0, 1, or 2 real solutions.

Assumptions
1. We are given the equation $10x + y = -x^2 + 2$ .
2. We need to determine the number of real solutions to this equation.

To find the number of real solutions, we can treat the equation as a quadratic equation in terms of $x$ by isolating $y$ .
$y = -x^2 - 10x + 2$

The equation $y = -x^2 - 10x + 2$ is a quadratic equation in the standard form $y = ax^2 + bx + c$ , where $a = -1$ , $b = -10$ , and $c = 2$ .

To determine the number of real solutions, we can use the discriminant of the quadratic equation, which is given by $\Delta = b^2 - 4ac$ .

Calculate the discriminant using the values of $a$ , $b$ , and $c$ .
$\Delta = (-10)^2 - 4(-1)(2)$

Simplify the discriminant.
$\Delta = 100 - (-8)$ $\Delta = 100 + 8$ $\Delta = 108$

Since the discriminant $\Delta > 0$ , the quadratic equation has two distinct real solutions.
The equation $10x + y = -x^2 + 2$ has two real solutions.