Answer:
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
x=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
6
2
−
5
−
4
=
0
6x^{2}-5x-4=0
6x2−5x−4=0
=
6
a={\color{#c92786}{6}}
a=6
=
−
5
b={\color{#e8710a}{-5}}
b=−5
=
−
4
c={\color{#129eaf}{-4}}
c=−4
=
−
(
−
5
)
±
(
−
5
)
2
−
4
⋅
6
(
−
4
)
√
2
⋅
6
Step-by-step explanation:
The answer is 20x-4, hope this helps.
A square matrix that is not invertible is called singular or degenerate.
31 divided by 9548 is 0.00324675
Theorem: If a function y = f(x) has a real root of b, then (x – b) is a factor of f(x).
As given in the problem, there are two roots: –2 and 1/2. The multiplicity of 1/2 is 2 meaning that the root 1/2 repeats twice. So the function f(x) can be written like this.
f(x) = k• (x – (–2))(x – 1/2)^2 = k•(x + 2)(x – 1/2)^2
We're supposed to find the coefficient k to complete the function.
Given that f(–3) = 5, we can plug –3 in for x and 5 in for f(x).
So 5 = k •(–3 + 2)(–3 – 1/2)^2
5 = k(–1)(–7/2)^2
5 = -k•49/4
Then 5 • 4/49 = -k
Or k = –20/49
So the function with the least degree is
f(x) = –20/49 (x + 2)(x – 1/2)^2.
