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nevsk [136]
4 years ago
9

How do you do the last question using the quadratic formula?

Mathematics
1 answer:
Anettt [7]4 years ago
3 0

Answer:

x = -1

Step-by-step explanation:

To do the last question using the quadratic formula, you first need the equation in standard form.

ax² + bx + c = 0.

To convert 2(x - 2)(x + 1) = x² - 4x - 5 into standard form, <u>simplify by expanding and collecting like terms. Then, have the equation equate to "0"</u> by moving everything to one side.

2(x - 2)(x + 1) = x² - 4x - 5        Expand brackets first using FOIL

2(x² + x - 2x - 2) = x² - 4x - 5        Collect like terms in brackets (x - 2x = -x)

2(x² - x - 2) = x² - 4x - 5        Distribute, multiply bracket numbers by "2"

2x² - 2x - 4 = x² - 4x - 5        Now make the equation equal 0

2x² - 2x - 4 - x² = x² - 4x - 5 - x²        Subtract x² from both sides

x² - 2x - 4 = -4x - 5        "x²" eliminated from the right side. Simplify left side.

x² - 2x - 4 + 4x = -4x - 5 + 4x        Add 4x to both sides.

x² + 2x - 4 = -5        "4x" eliminated from right side. Simplify left side.

x² + 2x - 4 + 5 = -5 + 5        Add 5 to both sides to eliminate it on the right.

x² + 2x + 1 = 0               Simplified left side.

This is now in standard form. State the "a", "b" and "c" values based on the standard form variables.

a = 1; b = 2; c = 1

<u>Substitute into the quadratic formula</u>

x = \frac{-b±\sqrt{b^{2}-4ac} }{2a}      (Please ignore the Â, it's a formatting error)

x = \frac{-2±\sqrt{2^{2}-4(1)(1)} }{2(1)}   Simplify the square root

x = \frac{-2±\sqrt{0} }{2}     The square root of 0 is 0.

x = \frac{-2}{2}      The numerator can only be -2. Simplify the fraction

x = -1         <u>Only one answer for "x"</u>.

Whenever the square root equals "0", there will only be one answer for "x".

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<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

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Answer:   g(x) = -(x + 1.25)² + 2.5625

<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

and input the given points for (x, y) to create a system of equations, then solve for a, h, and k.

EQ1: 2 = a(-2 - h)² + k  

       2 = a(4 + 4h + h²) + k

       2 - 4a - 4ah - ah² = k


EQ2: 1 = a(0 - h)² + k

        1 = ah² + k

        1 - ah² = k


EQ3: -2.5 = a(1 - h)² + k

        -2.5 = a(1 - 2h + h²) + k

        -2.5 -a + 2h + ah² = k

<u>Substitute</u> - set EQ1 = EQ2 and EQ2 = EQ3 to eliminate k

EQ1 = EQ2:     2 - 4a - 4ah - ah² = 1 - ah²

                                 1 - 4a - 4ah = 0    

EQ2 = EQ3:     1 - ah² = -2.5 - a + 2ah - ah²

                                3.5 + a - 2ah = 0  

<u>Elimination:</u>  - now solve the system for "a"

 1 - 4a - 4ah = 0    →    1(1 - 4a - 4ah = 0)    →     1 - 4a - 4ah = 0

3.5 + a - 2ah = 0   →   -2(3.5 + a - 2ah = 0) →  <u> -7 - 2a + 4ah = 0 </u>

                                                                         -6 - 6a           = 0

                                                                              -6a            = 6

<h2>                                                     a     = -1</h2>

Next, replace "a" with -1 into either of the equations to solve for "h"

                                         1 - 4a - 4ah = 0

                                     1 - 4(-1) - 4(-1)h = 0

                                            1 + 4 + 4h = 0

                                                 5 + 4h = 0

                                                       4h = -5

<h2>                                     h = -1.25</h2>

Now, replace "a" with -1 and "h" with -1.25 into any of the original equations (EQ1, EQ2, or EQ3) to solve for k:

1 - ah² = k

1 - (-1)(-1.25)² = k

1 - (-1)(1.5625) = k

1 + 1.5625 = k

<h2> 2.5625 = k</h2>

Now, input (h, k) = (-1.25, 2.5625) and a = -1 into the vertex formula:

y = -1(x - (-1.25))² + 2.5625    →     y = -(x + 1.25)² + 2.5625

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