Question

(A) x^2 + 6x + 9 = 0 (B) 8x^2+ 5x - 6 = 0 (C) (x + 4)^2 - 36 = 0

Answer 1

Answer:

Query (A)

${ \rm{ {x}^{2} + 6x + 9 = 0 }} \\ { \rm{(x + 3)(x - 1) = 0}} \\ { \boxed{ \rm{x = {}^{ - }3 \: \: and \: \: 1 }}}$

Query (B)

${ \rm{ {8x}^{2} + 5x - 6 = 0}} \\ \\ { \rm{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }} \\ \\ { \rm{x = \frac{ - 5 \pm \sqrt{217} }{16} }} \\ \\ { \boxed{ \rm{ \: x = 0.608 \: \: and \: \: {}^{ - } 1.233}}}$

Query (C)

${ \rm{ {(x + 4)}^{2} - 36 = 0 }} \\ \\ { \rm{ {(x + 4)}^{2} = 36 }} \\ \\ { \rm{x + 4 = \pm6}} \\ \\ { \boxed{ \rm{x = 2 \: \: and \: \: {}^{ - } 10}}}$

Answer 2

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

$x=\frac{-b +- \sqrt{b^2-4ac}}{2a}$

now substituting value:

$x=\frac{-5+-\sqrt{5^2-4*8*-6} }{2*8}$

$x=\frac{-5+-\sqrt{217}}{16}$

taking positive

$x=\frac{-5+\sqrt{217}}{16}$

taking negative

$x=\frac{-5-\sqrt{217}}{16}$

(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: