Question

medals & fans rewarded

Answer 1

what do you need help with?

Answer 2

\[7x ^{2}-35x=168\]

Answer 3

Divide by 7 first

Answer 4

what do you get?

Answer 5

-5x=24

Answer 6

7x^2 - 35x -168 = 0

ax^2 + bx + c = 0

x= [-b (+/-) sqrt(b^2 -4ac)]/(2a)

Answer 7

@hea i got -5x=24

Answer 8

Nope, \[7x^2-35x = 168\]when divided by 7 goes to \[x^2 - 5x = 24\]

Answer 9

ok

Answer 10

From there we can move the 24 across, giving us\[x^2 - 5x -24 = 0\]And then we can go about factorising. We are looking for 2 numbers. When added together they will make -5 and when multiplied together they will make -24

Answer 11

Can you think of any numbers that do that? (hint: one is positive and the other is negative)

Answer 12

3 & -8

Answer 13

Great! So when we factorise we get \[(x+3)(x-8) = 0\]Which is basically the solution because we can tell that clearly \(x = -3\) or \(x = 8\) Does that make sense?

Answer 14

ya it does thnx

Answer 15

you can either have 3 & -8 or 8 & -3
To check which one is true you should derivate the equation and solve for y=0 to see if the x is positive and negative. Then you would pick -3 & 8

Just use the resolvent formula. You won't make any mistake and its easier

Answer 16

\[7x^{2}- 35x-168=0\]
divide thru by 7,
\[x^{2}-5x-24=0\]
factorise,
\[7^{2}-8x+3x-24=0\]
\[x(x-8)+3(x-8)=0\]
\[(x-8)(x+3)=0\],
\[x-8=0\] OR \[x+3=0\]
Thus, x=8 or x=-3