COMPLETING THE SQUARE:
2x² + 7x + 3 = 0
Take the two common from the x² so
2 (x² + 7/2x + 3/2) = 0
Take the middle term (with x) which is (7/2) and half it (7/4) and now square it (7/4)² = 49/16 and now add and subtract this into the equation like this:
2 ( x² + 7/2x + 49/16 – 49/16 + 3/2) = 0
The first three terms in bracket make a perfect square which is (x + 7/4)² therfore
(x + 7/4)² – 49/16 + 3/2 = 0/2
(x + 7/4)² – 25/16 = 0
(x + 7/4)² = 25/16
By taking the square root on both sides this becomes:
x + 7/4 = ±?(25/16)
x + 7/4 = ±5/4
(i) x + 7/4 = 5/4 (ii) x + 7/4 = -5/4
(i) x = -1/2 (ii) -3 ANSWER.
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I’ll now try to solve this same equation using the quadratic formula and hope to produce the same result we got usign the ‘completing square’ method:
The quadratic formula is [-b ±?(b² - 4ac)] / 2a
In the equation 2x² + 7x + 3 = 0, a = 2, b = 7 and c = 3 and we will put these values into the quadratic formula and solve:
x = [-b ±?(b² - 4ac)] / 2a
x = [-7 ±?((7)² - 4(2)(3))] / 2(2)
x = [-7±?(49 - 24)] / 4
x = [-7±?25] / 4
x = (-7 ± 5)/4
(i) x = (-7 + 5)/4 AND (ii) x = (-7 – 5)/4
(i) x = -2/4 AND (ii) x = -12/4
x = -1/2 AND (ii) x = -3
Since both of these yield the same answer thefore it’s now your choice to use either one you would like.
I hope this helps.
Strictly speaking, the quadratic formula is derived from
the method of the completing the square.
Let ax^2+bx+c=0, then
x^2+(b/a)x+c/a=0 ( a is not 0)
x^2+(b/a)x+(b/2a)^2-(b/2a)^2+c/a=0
(x+b/2a)^2-[sqrt(b^2-4ac)/2a]^2=0=>
[x+b/2a+sqrt(b^2-4ac)/2a][x+b/2a-
sqrt(b^2-4ac)/2a]=0=>
x=[-b+/-sqrt(b^2-4ac)]/2a
x^2+4x+4=0
Solving by quadratic formula.
a=number in front of x^2 (1)
b=number in front of x term (4)
c=constant number with no x (4)
formula is X= -(b) +/- sqrt of [b^2-4(a)(c)] all divided by 2(a)
so x=[ -(4) +/- sqrt of {(4)^2-4(1)(4)}]/2(1)
x= [-4 +/- sqrt of (16-16)]/2
x=[-4 +/- sqrt of 0]/2
x=(-4+0)/2 and (-4-0)/2
x=-2; -2
so x=-2 with a multiplicity of 2 (same 2 answers for this one)