How do i solve the pair of simultaneous equations
y=4-2x
y=2x^2-3x+1 ??
how do i find minimum value of 2x^2-3x+1 and the value of x for which the minimum occurs???
Ah, simultaneous equations and calculus involved! Let me put on my clown nose and help you out with a dose of humor! 🤡
To solve the pair of simultaneous equations, we can substitute one equation into the other. Let's start with the first equation:
y = 4 - 2x
Now, substitute this value of y into the second equation:
4 - 2x = 2x^2 - 3x + 1
Rearrange this equation:
2x^2 - x - 3 = 0
Now, we have a quadratic equation! To find the minimum value of 2x^2 - 3x + 1, we need to first take the derivative with respect to x and set it equal to zero. But before we dive into the math, remember that clowns love to have fun, so let's keep things light and amusing! 🎪
Now, let's find the derivative of 2x^2 - 3x + 1:
d(2x^2 - 3x + 1)/dx = 4x - 3
To find the minimum value, we set the derivative equal to zero:
4x - 3 = 0
Solving this equation gives us:
x = 3/4
Now, substitute this value back into the original equation to find the minimum value:
2(3/4)^2 - 3(3/4) + 1 = -11/8
So, the minimum value of 2x^2 - 3x + 1 is -11/8. And the value of x for which the minimum occurs is 3/4.
Remember, math doesn't always have to be boring! 🤡
equate the two right sides, since they are both equal to y
so
2x^2 - 3x + 1 = 4 - 2x
2x^2 - x - 3 = 0
(2x-3)(x+1) = 0
2x-3=0 or x+1=0
x = 3/2 or x = -1
sub those values into y = 4-2x to get the corresponding y values of the points of intersection.
for the second part, are you taking Calculus.
If so, you should surely know how to do this. Take the first derivative, set it equal to zero and solve for x
Sub that back into the function to get the y.
If you are not taking Calculus you should know how to complete the square to find the vertex of the matching parabola.
Y = 4 - 2x
Y = 2x 2 - 3x + 1
To solve the pair of simultaneous equations, you can use the method of substitution or elimination. Let's use the method of substitution.
1. Start with the first equation: y = 4 - 2x.
2. Substitute this expression for y in the second equation: 4 - 2x = 2x^2 - 3x + 1.
3. Rearrange the equation to form a quadratic equation in standard form: 2x^2 - 3x - 2x + 4 - 1 = 0.
This simplifies to: 2x^2 - 5x + 3 = 0.
4. Factor the quadratic equation: (2x - 3)(x - 1) = 0.
Set each factor equal to zero and solve for x:
- 2x - 3 = 0 or x - 1 = 0.
This gives x = 3/2 or x = 1.
5. Substitute the values of x back into one of the original equations to find y.
For x = 3/2: y = 4 - 2(3/2) = 4 - 3 = 1.
For x = 1: y = 4 - 2(1) = 4 - 2 = 2.
So, the solutions to the pair of simultaneous equations are (3/2, 1) and (1, 2).
To find the minimum value of the quadratic equation 2x^2 - 3x + 1 and the value of x for which the minimum occurs, we can follow these steps:
1. To find the minimum value, we need to determine the vertex (the lowest point) of the parabolic graph represented by the quadratic equation.
2. The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c.
In this case, a = 2, b = -3, and c = 1.
Thus, x = -(-3) / 2(2) = 3/4.
3. Substitute the x-coordinate back into the quadratic equation to find the corresponding y-coordinate.
For x = 3/4: y = 2(3/4)^2 - 3(3/4) + 1 = 2(9/16) - 9/4 + 1 = 9/8 - 9/4 + 1 = 9/8 - 18/8 + 8/8 = 1/8.
So, the y-coordinate of the vertex is 1/8.
4. Therefore, the minimum value of the quadratic equation is 1/8, and the value of x where the minimum occurs is 3/4.
In summary, the minimum value of 2x^2 - 3x + 1 is 1/8, and the value of x for which the minimum occurs is 3/4.
To solve the pair of simultaneous equations given, you can follow the method of substitution.
1. Start by setting the two expressions for y equal to each other:
4 - 2x = 2x^2 - 3x + 1
2. Rearrange the equation to bring all terms to one side, setting it equal to zero:
2x^2 - 3x + 1 + 2x - 4 = 0
3. Combine like terms:
2x^2 - x - 3 = 0
4. To solve this quadratic equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -1, and c = -3. Substitute these values into the formula and solve for x.
x = (-(-1) ± √((-1)^2 - 4(2)(-3))) / (2(2))
= (1 ± √(1 + 24)) / 4
= (1 ± √25) / 4
5. Simplify the expression further:
x = (1 ± 5) / 4
This gives two possible solutions for x:
x1 = (1 + 5) / 4 = 6 / 4 = 3/2
x2 = (1 - 5) / 4 = -4 / 4 = -1
6. To find the corresponding values of y, substitute the values of x into either of the original equations. Let's use the first equation y = 4 - 2x:
For x1 = 3/2:
y1 = 4 - 2(3/2) = 4 - 3 = 1
For x2 = -1:
y2 = 4 - 2(-1) = 4 + 2 = 6
Therefore, the solutions to the pair of simultaneous equations are:
(x1, y1) = (3/2, 1)
(x2, y2) = (-1, 6)
Now, let's move on to finding the minimum value of the quadratic expression 2x^2 - 3x + 1, and the value of x for which the minimum occurs.
To find the minimum value, you can use calculus. Take the derivative of the quadratic equation with respect to x, and set it equal to zero to find the critical points.
1. Differentiate the quadratic equation:
f'(x) = 4x - 3
2. Set the derivative equal to zero and solve for x:
4x - 3 = 0
4x = 3
x = 3/4
This critical point represents a potential minimum or maximum for the function.
To determine if it is a minimum or maximum, you can use the second derivative test. Take the second derivative of the equation and evaluate it at the critical point.
1. Differentiate the derivative equation f'(x):
f''(x) = 4
2. Evaluate the second derivative at x = 3/4:
f''(3/4) = 4
Since the second derivative is positive at x = 3/4, it confirms that it is a minimum.
To find the minimum value, substitute the value of x = 3/4 into the quadratic equation:
f(3/4) = 2(3/4)^2 - 3(3/4) + 1
= 9/8 - 9/4 + 1
= 2/8
= 1/4
Therefore, the minimum value of the quadratic expression 2x^2 - 3x + 1 is 1/4, and the value of x for which the minimum occurs is x = 3/4.