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svp [43]
3 years ago
15

iddle" class="latex-formula">
how do i solve this with the pq- formula?
Mathematics
2 answers:
gayaneshka [121]3 years ago
8 0

Answer:

\displaystyle \boxed{\text{\sf \Large x=-1}}

Step-by-step explanation:

Add 1 to both sides

x^2+2x+1=0

Use quadratic formula with a=1, b=2, c=1

\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}

\displaystyle x= \frac{-(2) \pm \sqrt{(2)^2-4(1)(1)} }{2(1)}=-1

loris [4]3 years ago
6 0

Step-by-step explanation:

x²+2x+1

x²+x+x+1

x(x+1)1(x+1).

(x+1)(x+1)

x=-1. , x=-1

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Find the missing angles in the parallelogram.
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3 years ago
Can anyone help me solve this Algebra 2 Problem, i literally cant figure it out
Andrews [41]

Answer:

Zeros : 1 , -1, 3

Degree : 4

End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞

Y - intercept : -3

Extra Points: (0,-3), (2,-3)

Step-by-step explanation:

f(x) = 0 to find the zeros

Therefore (x+1)(x-1)^{2} (x-3) = 0

Clearly x = -1,1,3

Here 1 is a repeating root as it is (x-1)²

Degree is highest power of x in f(x)

Clearly it is x*x²*x = x⁴ is the maximum power of x

Thus degree is 4

Looking at end behavior we substitute x->∞ and x-> -∞

Clearly f(x)>0 as all terms are positive and f(x)->∞

Similarly when x->-∞

f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞

Y-Intercept is f(0)

f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3

Thus Y-Intercept is -3

Substitute x = 0 , 2 for extra points

Thus f(0) = -3

and f(2) = -3

Thus points on the graph (0,-3), (0,2)

We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.

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3 years ago
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Mekhanik [1.2K]
Function y = -2x + 5:
slope of -2
y intercept at 5
x intercept at 2 1/2

function y = x
slope of 1
y intercept at 0
x intercept at 0
8 0
2 years ago
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