How do you solve using the completing the square method x2+3x−10=0?

1 Answer
Tony B
Apr 13, 2016

yintercept=−10

Vertex→(x,y)→(−32,−494)

xintercepts=−5or+2

Explanation:

Given: x2+3x−10=0.......................(1)

Determine yintercept

Read directly from equation (1)

yintercept=−10

'~~~~~~~~~~~~~ Tip! ~~~~~~~~~~~~~~~~~~~~
As the equation is already in the form

y=a(x2+bax)+c

In this case a=1

xvertex=(−12)×ba=−32

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Step 1
Write equation (1) as (x2+3x)−10=0

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Step 2

Add the adjustment constant k
(x2+3x)−10+k=0

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Step 3
Move the power from x2 to outside the bracket
(x+3x)2−10+k=0
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Step 3

Remove the x from 3x
(x+3)2−10+k=0

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Step 4

Multiply the 3 inside the bracket by 12
(x+32)2−10+k=0
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Step 4
If we multiply out the bracket we end up with an additional term to those in the original equation. That term is (32)2 derived from (?+32)(?+32)=(32)2. This term must be removed which is achieved by making k=−(32)2

So now we have

(x+32)2−10+k=0→(x+32)2−10−(32)2=0

Completing the square →(x+32)2−494=0......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vertex→(x,y)→(−32,−494)

From equation (2)

(x+32)2=494

Square rooting both sides

x+32=±√49√4

xintercepts=−32±72=-5 or +2)#
Tony B

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