Question

finding zeros : f(x)=49x^2-144

how will i solve this one?

Answer 1

what can you tell about the numbers 49 and 144...?

Answer 2

perfect squares?

Answer 3

yep... so give you have 49x^2

this is the difference of 2 squares... thats why theres a subtraction sign

this may help you factorise

\[(ax +b)(ax - b)=a^2x^2 - b^2\]

after factoring let it equal zero...

then solve for x

Answer 4

are there any shortcuts on how to solve it?

Answer 5

well you know 49 and 144 are square number.... so what do you think a and b might be...?

Answer 6

7 and 12

Answer 7

or is it 7x+12 7x-12 ??

Answer 8

thats it so you are solving

(7x - 12)(7x + 12) = 0

so really you just need to solve

7x - 12 = 0

and

7x + 12 = 0

hope it makes sense

Answer 9

so the roots are -7 and 12???

Answer 10

no, you need to solve 7x - 12 = 0

then its

7x = 12

x = 12/7

thats one solution. You need to do the other

7x + 12 = 0

Answer 11

oohhhh

Answer 12

the solutions are the roots

Answer 13

7x=-12

x= -12/7

Answer 14

oh okay i understand it now thanks a lot... i have other problem here help me again please.. f(d)=2d^2-10d

Answer 15

i don't know how to solve it because it's not a perfect square number

Answer 16

Do you see a common factor?

Answer 17

2?

Answer 18

Yes, there is another one.

Answer 19

glad to help...

lots of good people here to help with the next question

Answer 20

2 and 2d?

Answer 21

u can use a identity here a^2-b^2=(a+b)(a-b)

Answer 22

$$\huge 2d^2-10d= 2d(d - 5)$$

Answer 23

But where's the other squared number?

Answer 24

I don't understand it :(

Answer 25

49x^2-144
can be written as (7x+12)(7x-12)
so x =12/7 and -12/7

Answer 26

these are the two roots

Answer 27

f(d)=2d^2-10d

Answer 28

That question was answered already @nirmalnema

Answer 29

Do you know how to do it now @MCath?

Answer 30

ohh...ok

Answer 31

i don't know this one f(d)=2d^2-10d