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Please solve x^2 + 6x - 7 = 0.  Please show your work.

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successyes | Valedictorian

Posted October 25, 2012 at 2:23 AM via web

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Please solve x^2 + 6x - 7 = 0.  Please show your work.

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3 Answers | Add Yours

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 25, 2012 at 5:07 AM (Answer #1)

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You may also complete the square as alternate method, using as pattern the formula `(a+b)^2 = a^2+2ab+b^2` .

You need to consider the first two terms of the equation such that:

`x^2 + 6x = x^2 + 2*3*x + 3^2 - 3^2`

Notice that you need to equate the mid term `2ab`  and `6x`  such that:

`2ab = 6x`

Since you already know `a = x` , then, you may substitute x for a such that:

`2xb = 6x => b = (6x)/(2x) = 3`

Since you need to add the square of 3 to complete the square, you also need to subtract `3^2` , to keep the balance such that:

`x^2 + 6x = (x+3)^2 - 9`

`x^2 + 6x - 7 = (x+3)^2 - 9 - 7`

Since `x^2 + 6x - 7 = 0` , then `(x+3)^2 - 9 - 7 = 0`  such that:

`(x+3)^2 - 9 - 7 = 0 => (x+3)^2=16 => x+3 = +-sqrt16`

`x+3 = +-4 => x_1 = -3+4 => x_1 = 1`

`x_2 = -3-4 => x_2 = -7`

Hence, evaluating the solutions to the given quadratic equation yields `x_1 = 1`  and `x_2 = -7` .

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najm1947 | Elementary School Teacher | Valedictorian

Posted October 25, 2012 at 3:42 AM (Answer #2)

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We have to solve x^2 + 6x - 7 = 0

This is in the form ax^2+bx+c=0 where a=1, b=6 and c=-7 and its roots are as:

x= [-b+sqrt(b^2-4ac)]/2a , [-b-sqrt(b^2-4ac)]/2a

Hence the roots are:

x = [-6+sqrt{(6)^2-4*1*(-7)}]/(2*1) and

x = [-6-sqrt{(6)^2-4*1*(-7)}]/(2*1)

=> x = [6+sqrt{36+28}]/2 , [6-sqrt{36+28}]/2

=> x = [6+sqrt{64}]/2 , [6-sqrt{64}]/2

=> x = [6+8]/2 , [6-8]/2

=> x = 14/2 , -2/2

=> x = 7 , -1

The solution of x^2 + 6x - 7 = 0 is x = 7, -1

 

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najm1947 | Elementary School Teacher | Valedictorian

Posted October 25, 2012 at 4:01 AM (Answer #3)

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Another method to solve x^2 + 6x - 7 = 0 is by factorisation as under:

x^2+6x-7 = 0 can be writen by bifercating 6x as under:

=> x^2 +7x - x - 7 = 0

=> (x^2 +7x) - (x+7) = 0

=> x*(x+7) -1*(x+7) = 0

=> (x+7)*(x-1) = 0

This show that either one or both the factors (x+7) and (x-1) are equal to zero.

x+7 = 0    =>   x = -7

x-1 = 0    =>    x = 1

Hence solution of x^2 + 6x - 7 = 0 is x = -7, 1

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Note: In the previous solution there is an error after:

x = [-6+sqrt{(6)^2-4*1*(-7)}]/(2*1) and

x = [-6-sqrt{(6)^2-4*1*(-7)}]/(2*1) and is corrected as under:

=> x = [-6+sqrt{36+28}]/2 , [-6-sqrt{36+28}]/2

=> x = [-6+sqrt{64}]/2 , [-6-sqrt{64}]/2

=> x = [-6+8]/2 , [-6-8]/2

=> x = 2/2 , -14/2

=> x = 1 , -7

The solution of x^2 + 6x - 7 = 0 is x = 1, -7

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