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(a+ 2) / (a + 1)  =  b/12

Rearrange as

12 ( a + 2) / ( a + 1)  =  b

This will converge  to ≈  12  as a becomes more positive and  more negative

a         b

0       24

1       18

2       16

3       15

5       14

11     13

-2       0

-3       6

-4       8

-5       9

-7      10

-13    11

x^2 -4x + 5 + x^2 -2x + 1 =  0

(x -1 ) (x + 5) + (x -1)^2  =  0

(x -1) [ ( x + 5) + (x -1) ]  = 0

(x -1) [ 2x + 4]  =  0

x -1  =  0           2x + 4  =  0

x = 1                 x =  -2

Sum  of the roots   =  -B /A  

-2 + 1 =  - B / A

-1  =  -B / A

1 = B/A

A = B

Product of  the Roots = C/A

-2 * 1  = C/A  =1/A

-2 = 1/A

-2A  = 1

A = -1/2

B = -1/2

A + B =    -1/2 + -1/2  =  -1

                                      a + ab =   250    

                                      a – ab = –240    

just add them               2a         =     10    

                                             a  =  5    

_.    

How many milliamps of current will I measure?

\(R=\frac{U}{I}=\frac{500V}{0.025A}\\ R=20k\Omega\)

The parallel-connected wire halves offer a quarter of the resistance of the original wire.

\(I=\frac{U}{R}=\frac{175V}{5k\Omega}=0,035A\\ \color{blue}I=35mA\)

We measure 35mA.

 !

Find the discriminant of the quadratic 5x^2 - 2x + 8 - 2x^2 + 6x + 2.     

When a quadratic is arranged in standard format, ax² + bx + c = 0,    

the discriminant is given by b² – 4ac.    

Combine like terms in the quadratic of interest and arrange in standard format.    

3x² + 4x + 10   

b² – 4ac    

4² – (4)(3)(10)    

16 – 120    

–104 is the discriminant    

_.    

\(f(x)=5x^2 - 2x + 8 - 2x^2 + 6x + 2\\ f(x)=3x^2+4x+10=0\\ \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ b^2-4ac=16-4\cdot 3\cdot 10=-104\\ (b^2-4ac)<0\)

The discriminant is less than zero. The function has no zero, no solution.

 !

Find the largest integer k such that the equation 5x^2 - kx + 8 - 2x^2 + 25 =0 has no real solutions  

Combine like terms and arrange equation    

in standard format ax² + bx + c = 0                3x² + (–k) x + 33 = 0    

The discriminant is defined as b² – 4ac          (–k)² – (4)(3)(33)    

                                                                          k² – 396    

An equation has no real solutions when the discriminant is negative.    

This is because in the quadratic formula, you calculate the square root of the discriminant.    

Therefore k² must be smaller than 396 to result in a negative discriminant.    

sqrt(396) is 19.9,  The next smaller integer is 19.  So, k = 19 answers the prolblem.         

_.    

Call AD  the altitude  =  x

tan B  =   AD / BD =  3/5

tan C = AD / CD =  1/4 =  3/12

BD =  (5)/(5 + 12) (32)  =  160 /17

CD = (12) / (5 +12) (32) = 384/17

So

tan B =  3/5  = AD / BD =  x / BD =    x / (160/17)   →  x = (3/5)(160/17)  = 96/17  =AD

Check this

tan C = 1/4 = AD /CD =  x / CD =  x / (384/17)  → x = (1/4) (384/17) =  96/17 = AD

[ABC] =  (1/2) BC *AD   = (1/2) 32 * (96/17)   = 1536 / 17

xy / (x + y)  =1           xz / (x + z) = 1

xy  = x + y                 xz = x + z

xy - y  = x                  xz - z  = x

y ( x -1) = x               z ( x -1) = x

y = x / (x -1)              z = x / ( x -1)

So   y = z

So

yz / (y + z)  = 1

z*z / ( z + z)  = 1

z^2  = 2z

z^2 - 2z  = 0

z ( z - 2) = 0

z= 0  reject      or     z = 2

Let D be the center of the red circle  and  r its radius

Let E be the center of the semi-circle with BC as a diameter

The radius of the large semi-circle = 2

We can form right triangle BDE with BD = 2-r , BE = 1 and the hypotenuse = ED = 1 + r

So

2^2  + (2-r)^2  = (1 + r)^2

4 + 4 - 4r + r^2  =  1 + 2r + r^2

7 = 6r

r = 7/6

sin A = 1/3 =  BD/ AB  = BD / 24  →  BD =  8

sin C =  1/2 =  BD / BC  = 8 / BC →  BC =  16

I create the function equation.

\(2x+86y+17=-4\\ 86y=-2x-21\\ \color{blue}f(x)=y=-\dfrac{1}{43}x-\dfrac{21}{86}\)

I determine the zero of the function.

\(f(x)=y=-\dfrac{1}{43}x-\dfrac{21}{86}=0\\ x=- \dfrac{21}{86}\cdot 43\\ \color{blue}x_0 =-10.5\)

 !

Find the largest value of $x$ such that $3x^2 + 17x + 15 = 2x^2 + 21x + 12 - 5x^2 + 17x + 34.$    

Given                               3x² + 17x + 15  =  2x² + 21x + 12 – 5x² + 17x + 34    

Combine like terms          6x² . . .   We can stop here; no need to bother with the other terms.    

The highest order term is the "squared term" and it is positive, which means that the curve    

is a parabola opening upward, for which there is no largest value of x.  It continues forever.    

We could say it approaches infinity, with the caveat that infinity can never be used as a value.    

_.   

y =  x^2 -8x + (8 + c)

This is aparabols that turns upward

The x coordinate of the vertex  =    8 / (2*1)    =  4

And we want the associated y coordinate of the vertex to   = 1

So

1 = (4)^2 - 8(4) + (8 + c)

1  =  -16 + 8 + c

9  = c

(2x + 7) (x - 4) + 31 - jx  = 0

2x^2 - ( 1 + j) x + 3 =  0

Thiswill have one real solutionwhen the discriminant  = 0

(1 + j)^2  - 4(2)(3)  =  0

(1 + j)^2  =  24        take  both roots

1 + j =  sqrt 24           and     1 + j  =-sqrt 24

j =  sqrt (24)  - 1                    j =  -sqrt (24) - 1

j = 2sqrt (6) - 1                     j = -2sqrt (6)  -1

Let E be the center of the large semi- circle

Note that the radius of this semi-circle is  = 2

So AB = EA = EB  = 2

So triangle AEB is equilateral  and its area  =  (1/2) 2^2 * sqrt (3)/. 2 =  sqrt (3)     (A)

And the  area between AB and  the edge of  the larger semi-circle =  pi * 2^2 * 60 / 360 =  (2/3)pi     (B)

So....the  white area of the small semicircle = B - A =   (2/3)pi -sqrt (3)

And the radius of the small semi-circle = 1

So the area of the small  semi-circle  = (1/2) pi (1)^2  =  pi /2

Shaded area =  pi/2 - ( 2/3 pi -sqrt 3) = sqrt (3) - pi/6

\(x^2 + 3y^2 = 18\\ 3y^2=18-x^2\\ y=\sqrt{6-\frac{x^2}{3}}\\ x=\sqrt{18-3y^2}\\ \color{blue}x+y=\sqrt{18-3y^2}+\sqrt{6-\frac{x^2}{3}}\)

 !

(x + 1) (x-2p)  =

x^2 + (1 -2p)x -2p

This is a parabola that turns upward

The x value of the vertex =  - (1 -2p) / (2) =  -1/2 + p 

We want this x value to produce  0...so....

(x + 1) ( x -2p)  = 0

( -1/2 + p + 1 ) ( -1/2 + p -2p)  = 0

(1/2 + p) ( -1/2 -p)  = 0

(1/2)+ p = 0               -1/2 - p = 0

p = -1/2                      p =-1/2

p = -1/2

The line has a slope = 5

The slope at any point on the parabola =  -12x

Set these = 

-12x = 5

x  = -5/12

So y= -6(-5/12)^2 =  -25/24

So

-25/24  = 5 (-5/12) + c

25/12 - 25/24 = c

50/24 - 25/24  =   c = 25/24

https://web2.0calc.com/questions/find-all-pairs-x-y-10-of-real-numbers-such-that-x

https://web2.0calc.com/questions/coordinates_11172

\( \dfrac{1}{\sqrt2+sqrt3}+\dfrac{1}{\sqrt2-sqrt3}. \)

[ sqrt 2 - sqrt 3 + sqrt 2 + sqrt 3]

__________________________ =

[sqrt 2 + sqrt 3] [ sqrt 2 -sqrt 3]

     2sqrt 2

____________  =

     2     -   3 

- 2sqrt 2