Can somebody please help me with these problems
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Can somebody please help me with these problems

[From: ] [author: ] [Date: 11-09-20] [Hit: ]
I hope this helps-what you need to do is factor the equation. then put each part to equal 0 and solve for it.......
x^2-4x+29=0

x2+3x-5=0

Can someone please explain to me how to do these? I'm so confused. It says to solve by completing the square. I have a test today and I am still clueless on how to do these types of problems..

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x^2 - 4x + 29 = 0

x^2 - 4x = - 29

complete square

(b/((2a))^2 = (4/((2)(1))^2 = (4/2)^2 = 2^2 = 4

x^2 - 4x + 4 = - 29 + 4

(x - 2)^2 = - 25

x - 2 = +- 5i

x = 2 +- 5i


x^2 + 3x - 5 = 0

This is of the form ax^2 + bx + c = 0

We have only the x^2 term and x terms on the left hand side. If you have an a that is not 1, you do not have to, divide through by it a but it makes completing the square easier and I would advice doing so. we do not need to here because you already have a = 1.

x^2 + 3x = 5

complete square

the b and a come from the coefficients of the x^2 and x terms

(b/((2)(a))^2 = (3/((2)(1))^2 = (3/2)^2 = 9/4

9/4 is what completes the square and we add it to both sides of the equation. always add to complete the square

x^2 + 3x + 9/4 = 20/4 + 9/4

I then factored the trinomial on the left hand side. If you did not notice it is equal to the 2nd to last step from completing the square. always use the sign of the existing x term in the paran below.

(x + 3/2)^2 = 29/4

took square root of both sides

x + 3/2 = +- sqrt(29)/2

subtracted 3/2 from both sides

x = - 3/2 +- sqrt(29)/2

I hope this helps

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what you need to do is factor the equation. then put each part to equal 0 and solve for it.
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