What is the solution of the linear-quadratic system of equations?

Question

{y=x^2+5x−3
{y−x=2

Answer 1

To solve the system of equations, we can substitute the expression for y from the second equation into the first equation:

y - x = 2
x^2 + 5x - 3 - x = 2
x^2 + 4x - 5 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula. Factoring it, we have:

(x+5)(x-1) = 0

Setting each factor equal to zero, we get:

x+5 = 0 or x-1 = 0

x = -5 or x = 1

Substituting these values back into the second equation, we can solve for y:

For x = -5:
y - (-5) = 2
y + 5 = 2
y = -3

For x = 1:
y - 1 = 2
y = 3

Therefore, the solution to the system of equations is x = -5, y = -3 and x = 1, y = 3.

Answer 2

What are the solutions of the quadratic equation?
2x^2 – 16x + 32 = 0

Answer 3

To find the solutions of the quadratic equation 2x^2 - 16x + 32 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a),

where a, b, and c are the coefficients of the quadratic equation.

In this case, the coefficients are a = 2, b = -16, and c = 32. Plugging these values into the quadratic formula, we have:

x = (-(-16) ± √((-16)^2 - 4(2)(32))) / (2(2))
x = (16 ± √(256 - 256)) / 4
x = (16 ± √0) / 4
x = (16 ± 0) / 4
x = 16/4
x = 4

Therefore, the solutions of the quadratic equation 2x^2 - 16x + 32 = 0 are x = 4.

Answer 4

Simplify the expression.

−5+i/2i

Answer 5

To simplify the expression (-5+i)/(2i), we can multiply the numerator and denominator by the conjugate of 2i, which is -2i. This will help eliminate the complex denominator.

(-5+i)/(2i) * (-2i)/(-2i)

Applying the distributive property, this becomes:
(10i-2i^2)/(-4i^2)

Now, let's simplify i^2, which is equal to -1:

(10i-2(-1))/(-4(-1))

Simplifying further:

(10i+2)/4

Dividing both terms by 2:

5i/2 + 1

The simplified expression is (5i/2) + 1.