Respuesta :
1.) a=5, b=9, c=-4
2.) C
3.) No solution
4.) x=-2, x=1.25
5.) x=-9, x=3
6.) quadratic formula
7.) 2
8.) 11cm by 16cm
100% for The Quadratic Formula and the Discriminant practice
2.) C
3.) No solution
4.) x=-2, x=1.25
5.) x=-9, x=3
6.) quadratic formula
7.) 2
8.) 11cm by 16cm
100% for The Quadratic Formula and the Discriminant practice
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:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]
[tex]x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-10)}}{2(4)} [/tex]
[tex]x = \frac{-3 \pm \sqrt{169}}{8} [/tex]
[tex]x_1 = \frac{-3 + 13}{8} [/tex]
[tex]x_1 = \frac{5}{4} [/tex]
[tex]x_2 = \frac{-3 - 13}{8} [/tex]
[tex]x_2 = -2 [/tex]
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
[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]
[tex]x = \frac{6 \pm \sqrt{6^2 - 4(-1)(27)}}{2(-1)} [/tex]
[tex]x = \frac{6 \pm \sqrt{144}}{-2} [/tex]
[tex]x_1 = \frac{6 + 12}{-2} [/tex]
[tex]x_1 = -9 [/tex]
[tex]x_2 = \frac{6 - 12}{-2} [/tex]
[tex]x_2 = 3 [/tex]
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
[tex]x = \frac{-27 \pm \sqrt{27^2 - 4(-1)(-176)}}{2(-1)} [/tex]
[tex]x = \frac{-27 \pm \sqrt{25}}{-2} [/tex]
[tex]x_1 = \frac{-27 + 5}{-2} [/tex]
[tex]x_1 = 11 [/tex]
[tex]x_2 = \frac{-27 - 5}{-2} [/tex]
[tex]x_2 = 16 [/tex]
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.
