Question

1. Which is the best method for solving the quadratic equation? Solve the quadratic equation using the method chosen. Leave all answers in simplest radical form.
Choose each method once.

• Take the square root of each side.

• Factor and use the zero product property.

• Complete the square.

• Use the quadratic formula.

A. x2 = –16x

B. y2 + 6y – 2 = 0

C. 2a2 = 72

D. p2 + 4p = 8


ANSWER:

























2. Consider the quadratic function y = –2x2 + 3x + 4.

A. Does the parabola open upward or downward? Explain.

B. Does the vertex lie on, below, or above the x-axis? Explain.


ANSWER:



















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Answer 1

Answer:

1. A. Factor and use the zero-product property; x = 0, -16

   B. Use the quadratic formula; y=-3-√11, -3+√11

   C. Take the square root of each side; x = -6, 6

   D. Complete the square; p= -2(√3 + 1). 2(√3 - 1)

2. A. Downward; coefficient of x² is negative

    B. Above; k is positive

Step-by-step explanation:

1. A. x² = –16x

Factor and use the zero-product property

$\begin{array}{rcl}x^{2} & = & -16x\\x^{2} + 16x & = & 0\\x(x + 16) & = &0\\x = \mathbf{0} & & x+ 16 = 0\\& & x = \mathbf{-16}\\\end{array}$

B. y² + 6y – 2 = 0

Use the quadratic formula

a = 1; b = 6; y = -2

$\begin{array}{rcl}y & = & \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\ & = & \dfrac{-6\pm\sqrt{6^2-4\times1\times(-2)}}{2\times1} \\\\ & = & \dfrac{-6\pm\sqrt{36+8}}{2} \\\\ & = & \dfrac{-6\pm\sqrt{44}}{2} \\\\ & = & \dfrac{-6\pm2\sqrt{11}}{2} \\\\ & = & -3\pm\sqrt{11}\\y=\mathbf{-3-\sqrt{11}} & &y= \mathbf{-3+\sqrt{11}}\\\end{array}$

C. 2a² = 72

Take the square root of each side.

$\begin{array}{rcl}2a^{2} & = & 72\\a^{2} & = & 36\\a & = & \pm 6\\a= \mathbf{-6} & & a = \mathbf{6}\\\end{array}$

D. p² + 4p = 8

Complete the square.

$\begin{array}{rcl}p^{2} + 4p & = & 8\\p^{2} + 4p + 4 & = & 12\\(p + 2)^{2}& = & 12\\p + 2& = & \pm \sqrt{12}\\& = & \pm 2\sqrt{3}\\p + 2 = -2\sqrt{3} & & p +2=-2\sqrt{3}\\p = -2 - 2\sqrt{3} & & p = -2 +2\sqrt{3}\\p= \mathbf{-2(\sqrt{3}+1)} & & p= \mathbf{2(\sqrt{3}-1)}\\\end{array}$

2. y = –2x² + 3x + 4

a = -2; b = 3; c = 4

A. Direction of opening

The parabola opens downward because the coefficient of x² is negative.

B. Vertex

The vertex form of a parabola is

y = a(x - h)² + k

where (h, k) are coordinates of the vertex.

The vertex will be above. on, or below the x-axis if k is positive, zero, or negative.

$\begin{array}{rcl}k& = & \dfrac{4ac-b^{2}}{2a}\\\\& = & \dfrac{4\times(-2) \times 4 - 3^{2}}{2\times4}\\\\& = & \dfrac{-32 - 9}{-8}\\\\& = & \dfrac{-41}{-8}\\\\& > &\mathbf{0}\\\end{array}$

The vertex is above the x-axis because k is positive.  

The graph below shows that your parabola opens downward and the vertex is above the x-axis.