Question

If (x+4)(x+1)-(x-1)(x-2)=0,what is the value of x?
plz answer ,I'll mark aa brainliest

Answer 1

a-b)^2=a^2+b^2–2ab

From the above formula below can be derived

(X-1/X)^2 =X^2 +(1/X)^2–2(X)*(1/X)

=(X+1/X)^2–4

(Derived from (a+b)^2=a^2 +b^2+2ab)

Now substitute the above derivation in the given equation

4((X+1/X)^2–4)-4(X+1/X)+1=0

Let X+1/X be t

Then

4(t^2–4)-4t+1=0

4t^2–16–4t+1=0

4t^2–4t-15=0

t^2–t-15/4=0

According to the Quadratic Formula, t , the solution for At2+Bt+C = 0 , where A, B and C are numbers, is given by

- B ± √ B2-4AC

t = ————————

2A

Where the value of A=1,B=-1,C=-15/4

On substituting the values as it is

You will be getting two values for t

5/2 and -3/2

But t=x+1/x

=> 5/2=x+1/x

=> x^2–5/2x+1=0

From the above formula

The roots of X will be

2 and 1/2

Hope this helps

Answer 2

I checked it it is correct you can also do it in your copy