Factor by completing the square 5x^2-21=10x

1 answer:

4 0

Solve for x over the real numbers:
5 x^2 - 21 = 10 x

Subtract 10 x from both sides:
5 x^2 - 10 x - 21 = 0

Divide both sides by 5:
x^2 - 2 x - 21/5 = 0

Add 21/5 to both sides:
x^2 - 2 x = 21/5

Add 1 to both sides:
x^2 - 2 x + 1 = 26/5

Write the left hand side as a square:
(x - 1)^2 = 26/5

Take the square root of both sides:
x - 1 = sqrt(26/5) or x - 1 = -sqrt(26/5)

Add 1 to both sides:
x = 1 + sqrt(26/5) or x - 1 = -sqrt(26/5)

Add 1 to both sides:
Answer: x = 1 + sqrt(26/5) or x = 1 - sqrt(26/5)

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Find the HCF using prime factors of 36 and 54

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Step-by-step explanation:

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HELP with these questions

zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)