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7(a squared)+32=7-40a 8(r squared)+3r+2=7(r squared) 42(x squared)-69x+20=7(x...

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aduitorhadrian | Student, Grade 11 | eNotes Newbie

Posted November 26, 2011 at 12:45 AM via web

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7(a squared)+32=7-40a

8(r squared)+3r+2=7(r squared)

42(x squared)-69x+20=7(x squared)-8

Maybe difficult - just warning you ahead of time!

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 26, 2011 at 1:15 AM (Answer #1)

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First equation is `7a^2 + 32 = 7 - 40 a.`

Consider the standard quadratic  equation: `ax^2 + bx + c = 0`

Make your equation to look just like the standard quadratic.

`7a^2 + 40a + 32 - 7 = 0 =gt 7a^2 + 40a + 25 = 0`

`` Use quadratic formula to find the roots.

`a_(1,2) = [-40+-sqrt(1600 - 700)]/14`

`` `a_(1,2) = (-40+-sqrt900)/14`

`a_(1,2) = (-40+-30)/14`

`a_1 = -10/14 =gt a_1 = -5/7`

`` `a_2 = -70/14 =gt a_2 = -5` 

 

The second equation is `8r^2 + 3r + 2 = 7r^2`

`` Subtract `7r^2 =gt 8r^2 - 7r^2+ 3r + 2 = 0`

`r^2+ 3r + 2 = 0`

`` `r_1 + r_2 = -` 3

`r_1*r_2 = 2 =gt r_1 = -1 and r_2 = -2`  

 

The next equation is `42x^2 - 69x + 20 = 7x^2 - 8`

Subtract `7x^2 - 8` :

`42x^2 - 69x + 20 - 7x^2+ 8 = 0`

`35x^2 - 69x + 28 = 0`

`x_(1,2) = [69+-sqrt(4761 - 3920)]/70`

`x_(1,2) = [69+-sqrt(841)]/70`

`` `x_(1,2) = (69+-29)/70`

`x_1 = 98/70 =gt x_1 = 14/10 = 7/5`

`` `x_2 = 40/70 =gt x_2 = 4/7`

ANSWER: `a_1 = -5/7`  and `a_2 = -5` ; `r_1 = -1`  and `r_2 = -2` ; `x_1 = 7/5 ` and `x_2 = 4/7` .

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beckden | High School Teacher | (Level 1) Educator

Posted November 26, 2011 at 1:42 AM (Answer #2)

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`7a^2 + 32 = 7 - 40a` Rewrite in standard form

`7a^2 + 40a + 25 = 0`

We need two numbers that multiplied together give 7*25 = 175 and add together to get 40

These numbers are 35 and 5

`7a^2 + 35a + 5a + 25 = 7a(a + 5) + 5(a + 5) = (7a + 5)(a + 5) = 0`

So either 7a+5 = 0 or a + 5 = 0.   Solving we get `a = -5/7` or `a = -5`

Second equation

`8r^2 + 3r + 2 = 7r^2` Rewrite in standard form

`r^2 + 3r + 2 = 0`

We need two numbers whose product is 2 and sum is 3, 2 and 1.

`r^2 + r + 2r + 2 = r(r + 1) + 2(r + 1) = (r + 2)(r + 1) = 0`

Using the zero product property r = -2 or r = -1

Third equation

`42x^2 - 69x + 20 = 7x^2 - 8` Rewrite in standard form

`35x^2 - 69x + 28 = 0`

We need two numbers whose product is 35*28 = 4*5*7*7 and sum is -69.

The numbers are -49 and -20.

`35x^2 - 49x - 20x + 28 = 0`

`7x(5x - 7) - 4(5x - 7) = 0`

`(7x - 4)(5x - 7) = 0`

Using the zero product property 7x - 4 = 0 or 5x - 7 = 0

We get `x = 4/7` or `x = 7/5`

So the answers are

`a^2+32 = 7 - 40a` solution is ` ` `a = -5` or `a = -5/7`

`8r^2 + 3r + 2 = 7r^2` solution is r = -2 or r = -1

`42x^2 - 69x + 20 = 7x^2 - 8` solution is `x = 4/7` or `x = 7/5`

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