Question

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

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

A) No solution
B) One real solution
C) Two real solutions
D) Two complex solutions

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

A) x = 4, x = 8
B) x = 6, x = 10
C) x = -2, x = 12
D) x = 16, x = 64

Simplify the expression:
-5 + i / 2i

A) -5 - i/2
B) -5 + i/2
C) 5 - i/2
D) 5 + i/2

Answer 1

The solution of the linear-quadratic system
`y = x^2 + 5x - 3` and y - x = 2 has two real solutions: (-5, -3) and (1, 3). The quadratic equation
`2x^2 - 16x + 32 = 0` has one repeated real solution: x = 4. To simplify the expression -5 + i / 2i, multiply by the conjugate to get -5 + i/2.

Step-by-step explanation:

The solution of the linear-quadratic system of equations
`y = x^2 + 5x - 3` and y - x = 2 is determined by substitution or elimination. For the first equation by substituting y from the second into the first, you get
`x^2 + 5x - 3 = x + 2`, which simplifies to
`x^2 + 4x - 5 = 0`. This can be factored to (x + 5)(x - 1) = 0, giving solutions x = -5 or x = 1.

Plugging these back into y - x = 2 gives the corresponding y values as y = -3 and y = 3, respectively. Therefore, the system has two real solutions: (-5, -3) and (1, 3), which corresponds to option C.

For the second part, the quadratic equation
`2x^2 - 16x + 32 = 0` can be simplified by dividing each term by 2, yielding
`x^2 - 8x + 16 = 0`, which is a perfect square trinomial.

Factoring gives (x - 4)(x - 4) = 0, so the only solution is x = 4. However, since the question suggests there are two values, there might be a misunderstanding or typo. In canonical form, if we perceive the quadratic as
`2(x - 4)^2 = 0`, the answer is still x = 4, which is repeated, and option A is the closest to this result.

To simplify the complex expression, -5 + i / 2i, multiply the numerator and denominator by the conjugate of the denominator, in this case, -2i.

It becomes (-5 + i)(-2i) / (2i)(-2i) = (10i - 2) / -4 = -5i/2 + 1/2, which corresponds to option B, -5 + i/2.