Question

MEDAL given work two problems out pleaseee 1. x^2- 8x -7=0 2. x^2 -3x=40

Answer 1

well if you are solving them

for question 1, you'll need the general formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

and a = 1, b = -8 and c = -7

just substitute and evaluate

Question 2.

rewrite the equation

\[x^2 - 3x - 40 = 0\]

this can be solved by factoring

what are the factors of -40 that add to -3.... the larger factor is negative

then the factored form is

(x + factor 1)(x + factor 2) = 0

then you need to solve

x + factor 1 = 0

x + factor 2 = 0

hope it helps.

Answer 2

@surjithayer can you help with two problems pleaseeee

Answer 3

1.
x^2-8x-7=0
x=[-(-8)+-sqrt{(-8)^2-4*1*-7}]/2*1
={8+-sqrt92}/2
=(8+-2sqrt23)/2
=4+-sqrt23
Hence x=4+sqrt23
and x=4-sqrt23

Answer 4

lol... why give answers... how does it help understanding...

Answer 5

2.
x^2-3x-40=0
1*-40=-40
8-5=3
8*-5=-40
write -3x=-(8-5)x then make factors.

Answer 6

@surjithayer you did the good like always thanks!

Answer 7

@campbell_st actually i have the answer i just need to work the problem out so whatever!

Answer 8

lol... so why not post your answers with the questions... it makes the process of checking so much easier...

Answer 9

and a very simple way to check you solutions is to substitute back into your equation... and see if you get the desired answer... zero...

Answer 10

thats true thankssss @campbell_st