How do you solve 2x^2-5x-3=02x2−5x−3=0 by completing the square?

2 Answers
Roy Ø.
Jun 6, 2018

x_1=3x1=3, x_2=-1/2x2=−12

Explanation:

Completing the square means that you want to write 2x^2-5x-3=02x2−5x−3=0
on the form (x-a)^2=b(x−a)2=b

First, let's get rid of the constant in front of the 2nd degree term:
x^2-5/2x-3/2=0x2−52x−32=0
We want this on the form
(x-a)^2=b(x−a)2=b
i.e. x^2-2ax+a^2=bx2−2ax+a2=b
Comparing term for term, we see that
2a=5/22a=52, i.e. a=5/4a=54
(x-5/4)^2=x^2-5/2x+25/16(x−54)2=x2−52x+2516
We, therefore, have
(x^2-5/2x+25/16)-25/16-3/2=0(x2−52x+2516)−2516−32=0
or (x^2-5/2x+25/16)=25/16+24/16=49/16(x2−52x+2516)=2516+2416=4916
(x-5/4)^2=(7/4)^2(x−54)2=(74)2
x-5/4=+-7/4x−54=±74
x=5/4+-7/4x=54±74
Therefore x_1= 5/4+7/4=3x1=54+74=3
x_2=5/4-7/4=-1/2x2=54−74=−12

Sardar ji · Stefan V.
Jun 6, 2018

x = 3x=3 or x=-1/2x=−12

Explanation:

Given:

2x^2 -5x -3 =02x2−5x−3=0

On dividing this equatin by 22, we get:

x^2 -5/2x -3/2 =0x2−52x−32=0

x^2 -5/2x = 3/2x2−52x=32

Add (5/4)^2(54)2 on both sides

x^2 -5/2x +(5/4)^2 = 3/2 + (5/4)^2x2−52x+(54)2=32+(54)2

[because a^2 + b^2 + 2ab = (a+b)^2a2+b2+2ab=(a+b)2 ]

So

(x-5/4)^2 = 3/2 + 25/16(x−54)2=32+2516

(x-5/4)^2 = 49/16(x−54)2=4916

Taking the square root on both sides and solving it.

x-5/4 = 7/4" "x−54=74 or " "x-5/4 = -7/4 x−54=−74

x = 7/4+5/4" "x=74+54 or " "x = -7/4+5/4 x=−74+54

x = 12/4" "x=124 or " "x = -2/4 x=−24

x = 3" "x=3 or " "x = -1/2 x=−12

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