I wrote this like three times and i still got no answer, besides a wrong one? please help?

Question

7k^2+10k-100=2k^2+55 Please help solve it.?
Can you show me how to do it in fractions step by step please i want to learn.
Explain it to me like if i was a retard please.
completing the square

i will really appreciate if u help me out please

Answer 1

group all variables on one side (variables are the letters or alphabets in the equation).
7k^2-2k^2+10k=55+100
5k^2+10k=155 (you can't add the "10k" to "5k^2" because they are not the same,the exponent makes the difference.)
bring the variables and numbers on the same side and equate to zero.Now we have a quadratic equation.
5k^2+10k-155=0 (lets divide thru by 5 since 5 is the least common factor.note that it isn't necessary to do so but it will make the values look smaller).
k^2+2k-31=0
the general quadratic equation (-b+/- square root of(-b^2-4ac))/2a
a= 1 from k^2,b=2 from 2k and c= -31
(-2+/- sqrt(-2^2-4(1x-31))/2(1)
(-2+/- sqrt(4+124))/2
(-2+/- sqrt (128))/2 ,
(-2+ 11.31)/2=4.655~5.0 or
(-2-11.31)/2= -6.655~ -7.0
Now you can plugin the two answers the one that fits is the most correct.i solved for negative and positive because a square root is both positive and negative.

Answer 2

This is a little complex; ill try and simplify it:

first rearrange the equation so you have everything on one side:

therefore:

(7k^2 - 2k^2) + 10k + (-100 - 55) = 0
5k^2 + 10k - 155 = 0

Now take out 5 as a common factor:

5 (k^2 + 2k - 31) = 0

Divide both sides by 5 to get rid of it:

k^2 + 2k - 31 = 0

Now comes completing the square:

Look at the middle term; 2k

Now, the expanded general form of a perfect square is:

(k + x) = x^2 + 2xk + k^2

compare this to the simplified eqaution above:

We need to find a value of x which satisifies the general eqaution of a perfect sqaure.

Therefore:

2xk = 2k

Divide both sides by 2k:

x = 1

So, the number which must be added and subtracted is (1^2) = 1

therefore

(k^2 + 2k + 1) - 1 -31 = 0

Factorise the bit in brackets:

(k + 1)^2 - 32 = 0

Then solve with algebra:

(k + 1)^2 = 32

Rember, when taking a sqaure root, there are two amswers; plus and minus:

(Note: { represents sqaure root)

k + 1 = +- {32}
k = +- {32} - 1

Now simplify the surd:

k = +- {16 * 2} - 1
k = +- 4{2} - 1

Hope it helps.

Answer 3

Start with 7k^2+10k-100=2k^2+55

Subtract 2k^2+55 from both sides to get:

5k^2 + 10k - 155 = 0

Since all the coeficients are divisible by 5, divide through by 5 to make it a bit simpler.

k^2 + 2k - 31 = 0

Now you can solve with the quadratic formula or complete the square (note that the quadratice formula is essentially completes the square, but lets you skip actually doing that).

To complete the square you want to divide the coefficient of the k term by 2 and then square it to get 1, then add and subtract 1 to get:

k^2 + 2k + 1 - 32 = 0

The k^2 + 2k + 1 now factors into (k+1)^2. Notice that you got the 1 when you divided the coefficient of the k term in half. So you have:

(k+1)^2 - 32 = 0

Add 32 to both sides to get

(k+1)^2 = 32

Taking the square root of both sides gives you k+1 = plus or minus the square root of 32. So you have two solutions:

k = -1 + squareroot(32), -1 - squareroot(32)

Hope that helps!

Answer 4

OK - to solve by completing the square.

First get all the k^2 and k terms on the left and other numbers on the right.

5 k^2 + 10k = 155 Divide both sides by 5.
k^2 + 2k = 31 Now we complete the square on the left.

Since there are 2k, divide 2 in half (= 1), and square that number to get the number that will complete the square (1^2 = 1)

k^2 + 2k + 1 = 31 + 1 We had to add 1 to the other side to keep it even.

(k + 1) (k + 1) = 32 Now take the square root of both sides.
k + 1 = plus or minus the square root of 32
k + 1 = plus or minus the square root of 16*2
k + 1 = 4 * plus or minus square root of 2
k = 4 * plus or minus square root of 2 minus 1

Answer 5

(7-2)k^2 +10k -100-55=0

5k^2 +10k -155 = 0 2nd degree equation

k=((-10 +- sqrt(100 +20*55))/10 applying a known formula