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Answer:
Query (A)
[tex]{ \rm{ {x}^{2} + 6x + 9 = 0 }} \\ { \rm{(x + 3)(x - 1) = 0}} \\ { \boxed{ \rm{x = {}^{ - }3 \: \: and \: \: 1 }}}[/tex]
Query (B)
[tex]{ \rm{ {8x}^{2} + 5x - 6 = 0}} \\ \\ { \rm{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }} \\ \\ { \rm{x = \frac{ - 5 \pm \sqrt{217} }{16} }} \\ \\ { \boxed{ \rm{ \: x = 0.608 \: \: and \: \: {}^{ - } 1.233}}}[/tex]
Query (C)
[tex]{ \rm{ {(x + 4)}^{2} - 36 = 0 }} \\ \\ { \rm{ {(x + 4)}^{2} = 36 }} \\ \\ { \rm{x + 4 = \pm6}} \\ \\ { \boxed{ \rm{x = 2 \: \: and \: \: {}^{ - } 10}}}[/tex]
Answer:
solution given:
(A) x^2 + 6x + 9 = 0
doing middle term
x^2+ (3+3)x+9=0
x^2 +3x+3x +9=0
taking common from two each term.
x(x+3)+3(x+3)=0
(x+3)(x+3)=0
either
x+3=0
x=-3
(B) 8x^2+ 5x - 6 = 0
Comparing above equation with
ax^2+bx+c=0,
we get,
a=8
b=5
c=-6
now
we have
[tex]x=\frac{-b +- \sqrt{b^2-4ac}}{2a}[/tex]
now substituting value:
[tex]x=\frac{-5+-\sqrt{5^2-4*8*-6} }{2*8}[/tex]
[tex]x=\frac{-5+-\sqrt{217}}{16}[/tex]
taking positive
[tex]x=\frac{-5+\sqrt{217}}{16}[/tex]
taking negative
[tex]x=\frac{-5-\sqrt{217}}{16}[/tex]
(C) (x + 4)^2 - 36 = 0
(x + 4)^2 - 6^2 = 0
it is in the form of x²+y²:(x+y)(x-y)
so (x + 4)^2 - 6^2 can be written as (x+4+6)(x+4-6)
above equation becomes
(x+10)(x-2)=0
either
x=-10
or
x=2
Step-by-step explanation: