The Quadratic Equation c² + 6c+ 8 = 0 Show work please

Mathematics

2 answers:

vaieri [72.5K]3 years ago

8 0

The answers c = -2 and c = -4 would solve the given equation.
hope it helps u

Bad White [126]3 years ago

6 0

The formula for the quadratic formula is x (c in this case) = (-b(+/-)√(b²-4ac))/2a
This is used for an equation in standard quadratic form: ax² + bx + c = 0
1.) Put it in the correct form, if not already in it.
Ex. c² + 6c + 8 = 0
2.) Identify each part of the equation:
a = 1 (the leading coefficient), b = 6 (the coefficient in front of the second variable), c = 8
3.) Plug in each variable answer
c = (-6(+/-)√(6²-4(1)(8))/2(1)
4.) Simplify
c = (-6(+/-)√(36-(4*8))/2
c = (-6(+/-)√(36-32))/2
c = (-6(+/-)√(4))/2
c = (-6(+/-)2)/2
*Here, the equation splits in two. It becomes:
c = (-6+2)/2 AND c = (-6-2)/2
*Simplify again:
c = -4/2 AND c = -8/2
c = -2 AND c = -4
The answers c = -2 and c = -4 would solve the given equation.
Hope this helps! :)

You might be interested in

What is the surface area of this design?

klasskru [66]

I think it is 7200in.

3 0

3 years ago

Read 2 more answers

2 if x = 137,2° and sin(x-y) Determine the numerical value of: cos y.tanx y = 114,99

Amanda [17]

Answer:

....................

3 0

3 years ago

Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .