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Bas_tet [7]
4 years ago
13

Complete the square of 4 x squared plus 8X minus 5 equals 0

Mathematics
2 answers:
n200080 [17]4 years ago
6 0
The answer to your question above  is 24x-5
Mice21 [21]4 years ago
6 0
First, subtract $-5$ from both sides:
$4{x}^{2}+8x=5$
Divide both sides by 4:
${x}^{2}+2x=\frac{5}{4}$
Add $({\frac{b}{2}})^{2}=({\frac{2}{2}})^{2}=1$ to both sides to complete the square:
${x}^{2}+2x+1=\frac{9}{4}
${(x+1)}^{2}=\frac{9}{4}$
We now take the square root of both sides, getting two different equations:
$x+1=\frac{3}{2}$ or $x+1=-\frac{3}{2}$
Solving each, we get $\left \{ {{x_1=\frac{1}{2}} \atop {x_2=-\frac{5}{2}} \right. $
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Read 2 more answers
Find the area of the following regions, expressing your results in terms of the positive integer n ≥ 2. The region bounded by
cluponka [151]

Answer:

The area of the searched region is A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2

Step-by-step explanation:

If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.

In this case, for 0<x<1, f(x)<g(x)

while for 1<x, g(x)<f(x).

Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:

A=A(a<x<1)+A(1<x<b)

A(a

A(1

A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} +  b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1}  - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}   -2

3 0
3 years ago
1.- (A^2-6a+4)/(2a)<br> 2.- (2b^2+b-8)/(2b)
siniylev [52]

Answer:

ok

Step-by-step explanation:

7 0
4 years ago
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