1.) a=5, b=9, c=-42.) C3.) No solution 4.) x=-2, x=1.255.) x=-9, x=36.) quadratic formula 7.) 28.) 11cm by 16cm100% for The Quadratic Formula and also the Discriminant practice

the actual solution: x = - 2

Find the attached document for the staying solution

Step-by-step explanation:

The equation given is:

x3 - 5x2 + 28 = 0 

Let assume that -2 is among the source of the equation. Substitute -2 for x

(-2)^3 - 5(-2)^2 + 28

-8 - 20 + 28 = 0

Therefore, -2 is one of the root of the equation because the equation have tendency to zero.

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If x = -2, climate x+2 is just one of the components of the equation. Therefore, the genuine solution is x = -2

Please discover the attached document for the remaining solution.

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1. The equation 5x^2+9x=4, can be rewritten as 5x^2+9x-4=0, wherein a=5 (the coefficient the the quadratic term), b = 9 (the coefficient of the direct term) and also c = -4 (the live independence term).

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3. To usage 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 actual solution.

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

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5. Rewriting the formula, so the all the terms space on one side and also ordering the polynomial expression, gives: -x2-6x+27=0. Right here a = -1, b = -6 and c = 27. Utilizing the quadratic formula gives

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

7. When the discriminant is better than zero, there are two different solutions and also when the is same to zero, it has one dual solution.

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

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

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

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

Isolating a in equation 1, provides (units room omitted):

2a + 2b = 54

2a = 54 - 2b

a = 27 - b (eq. 3)

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

(a)(b) = 176

(27 - b)(b) = 176

-b^2 + 27b - 176 = 0

Using quadratic formula, we get

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Then, there room two possible values because that b, 11 cm and also 16 cm. Replacing this values right into equation 3 gives: