Question

answer the stand form of (3x-5) (3x+5) = 10
(x-3)quantity squared + 7 =0
2(x-1)quantity squared -6 =0
x(2x+7=(x+3) (x-3)
-3x(x+5) = 8
(2x+3) quantity squared (x+3) quantitysquared

.. JESA :)

Answer 1

Is it 10(x-3) ^2..? Or the whole equation squared?

Answer 2

Please writr properly using equation.

Answer 3

what ?

Answer 4

Is it (3x-5) (3x+5) = 10(x-3)^2+ 7= 0 ?

Answer 5

yes .. wat is the answer ?

Answer 6

Solve (3x-5)(3x+5)=0

Answer 7

9x_25=0 .. ?

Answer 8

yes jesa it is : \((3x-5)(3x+5)=0\) ===>\(9x^2-25\)

Answer 9

how the others .. ?

Answer 10

try to do them on ur own .. tell me where r u having problem

Answer 11

plzzzzzzzz .

Answer 12

uhhhhmm .. i dont know how to solve it !

Answer 13

im sorry !

Answer 14

can u write the second one in expression form ?

Answer 15

i mean : (x-3)quantity squared + 7 =0
this into an expression

Answer 16

did u get what i mean ?

Answer 17

(x-3)2 + 7+=0

Answer 18

ok very good ; \((x-3)^2+7=0\)

Do you know how to solve \((x-3)^2\) ?

Answer 19

x squared-9 /

Answer 20

no .. @jesa


just try to find it out with this identity : \(\huge{(a-b)^2=a^2+b^2-2ab}\)

Answer 21

i dont know !

Answer 22

put a = x , b = 3

can u do it now ?

Answer 23

Okay Jesa.. \[(a-b)^{2}= (a-b)(a-b)\].. if you expnad you'll get \[a.a - a.b-b.a + b.b\], Which is equal to \[ a^{2}-2ab- b^{2}\]. Now can you solve?