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3 years ago

The solution to the equation 2x^2+x-1=2

Mathematics

1 answer:

maxonik [38]3 years ago

5 0

Answer:

$x=-\frac{3}{2}, x=1$

Step-by-step explanation:

$2x^2+x-1=2$

Subtract 2 from both sides and factor:

$2x^2+x-1=2\\2x^2+x-3=0\\(2x+3)(x-1)=0$

Solve each factor individually:

$2x+3=0\\2x=-3\\x=-\frac{3}{2}$

$x-1=0\\x=1$

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Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5i<br>a. 29i<br>b. 291-20<br>C.-21i<br>d.29

attashe74 [19]

Answer:

A. 29i

Step-by-step explanation:

Step 1: Plug in given variables

(2 - 5i)(2 + 5i)i

Step 2: Difference of squares (expand)

(4 - 25i²)i

Step 3: Imaginary numbers rules

(4 - 25(-1))i

Step 4: Combine like terms

(4 + 25)i

(29)i

Your final answer will be 29i

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arsen [322]

Congruent is ur answer

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3 years ago

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

BaLLatris [955]

Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Step-by-step explanation:

Given

$f(x) = x^2 - x +2$

Required

The vertex

We have:

$f(x) = x^2 - x +2$

First, we express the equation as:

$f(x) = a(x - h)^2 +k$

Where

$Vertex = (h,k)$

So, we have:

$f(x) = x^2 - x +2$

--------------------------------------------

Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

--------------------------------------------

$f(x) = x^2 - x +2$

$f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2$

$f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}$

Expand

$f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}$

Factorize

$f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}$

Factor out x - 1/2

$f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}$

Compare to: $f(x) = a(x - h)^2 +k$

$h = \frac{1}{2}$

$k = \frac{7}{4}$

Hence:

$Vertex = (\frac{1}{2},\frac{7}{4})$

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Step-by-step explanation:

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