Mathematics, 29.07.2019 02:00 bombo99

5x2=5x^2 4x2+4x^2 1. what values of a, b, and c would you use in the quadratic formula for the following equation? 5x2+9x=4 2. attachment 3. what are the approximate solutions of 2x2-x+10=0? 4. solve the equation using the quadric formula. 4x2+3x-10=0 5. solve the equation using the quadric formula. -x2+27=6x 6. which method is the best method for solving the equation? 8x2-13x+3=0 7. how many solutions are there for 5x2+7x-4=0? 8. the perimeter of a rectangle is 54 cm. the area of the same rectangle is 176 cm2. what are the dimensions of the rectangle?

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Answer from: zachcochran2007

1. The equation 5x^2+9x=4, can be rewritten as 5x^2+9x-4=0, where a=5 (the coefficient of the quadratic term), b = 9 (the coefficient of the linear term) and c = -4 (the independent term).

2. No attachment is shown.

3. To use the quadratic formula, the coefficients are: a = 2, b = -1 and c = 10. However, the discriminant, i. e., b^2 -4(a)(c) is negative, therefore, there is no real solution.

4. You have to use the quadratic formula, with a = 4, b = 3 and c = -10, as follows:

$x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}$

$x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-10)}}{2(4)}$

$x = \frac{-3 \pm \sqrt{169}}{8}$

$x_1 = \frac{-3 + 13}{8}$

$x_1 = \frac{5}{4}$

$x_2 = \frac{-3 - 13}{8}$

x_2 = -2

5. Rewriting the formula, so that all the terms are on one side and ordering the polynomial expression, gives: -x2-6x+27=0. Here a = -1, b = -6 and c = 27. Using the quadratic formula gives

$x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}$

$x = \frac{6 \pm \sqrt{6^2 - 4(-1)(27)}}{2(-1)}$

$x = \frac{6 \pm \sqrt{144}}{-2}$

$x_1 = \frac{6 + 12}{-2}$

x_1 = -9

$x_2 = \frac{6 - 12}{-2}$

x_2 = 3

6. When the second grade polynomial has all of its coefficients (a, b and c) different than zero, like this case, the best method for solving the equation is the quadratic formula .

7. When the discriminant is greater than zero, there are two different solutions and when it is equal to zero, it has one double solution.

The discriminant for this case is 7^2 - 4(5)(-4) = 129. So, there are two different solutions .

8. Let's call the sides of the rectangle a and b.

perimeter formula: 2a + 2b = 54 cm (eq. 1)

area formula: (a)(b) = 176 cm2 (eq. 2)

Isolating a in equation 1, gives (units are omitted):

2a + 2b = 54

2a = 54 - 2b

a = 27 - b (eq. 3)

Replacing equation 3 into equation 2, gives (units are omitted):

(a)(b) = 176

(27 - b)(b) = 176

-b^2 + 27b - 176 = 0

Using quadratic formula, we get

$x = \frac{-27 \pm \sqrt{27^2 - 4(-1)(-176)}}{2(-1)}$

$x = \frac{-27 \pm \sqrt{25}}{-2}$

$x_1 = \frac{-27 + 5}{-2}$

x_1 = 11

$x_2 = \frac{-27 - 5}{-2}$

x_2 = 16

Then, there are two possible values for b, 11 cm and 16 cm. Replacing these values into equation 3 gives:

a = 27 - 11 = 16 cm

a = 27 - 16 = 11 cm

Therefore, the two sides are 11 cm and 16 cm long, despite their names.

Answer from: Quest

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Answer from: Quest

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