Question

x2 - 2x - a (a+2) solve this equation answer is a+2 ,a
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Answer 1

Answer:

Before finding the roots of a quadratic equation, just find the determinant given by

det=b^2-4*a*c.

For this problem, det=4+8=12>0

So the roots are real and distinct.

So the roots would be

x=(-b+(det)^(1/2))/(2*a)

x=(-b-(det)^(1/2))/(2*a)

So x=(-2+2*(3^(1/2)))/2

x=(-2-2*(3^(1/2)))/2

the answer will be 1+/-3^1/2

How do I solve X^2+2X-2=0 by Quadratic formula?

Let a quadratic equation be

ax2+bx+c=0....(1)

What this actually means is that, at some value of x, (1) will be zero. That's the solution to equation.

Let's divide equation by a. (We can do that because a is non zero or else it won't be a quadratic equation).

So we get

x2+bax+ca=0=>x2+2xb2a+ca+(b2a)2−(b2a)2=0

If you know, (m+n)^2 = (m^2+2mn+n^2), then you can rewrite above equation as,

=>(x+b2a)2+ca−(b2a)2=0=>(x+b2a)2−((b2a)2−ca)=0=>(x+b2a)2−((b2a)2−ca−−−−−−−−√)2=0=>(x+b2a)2−(b2−4ac4a2−−−−−√)2=0=>(x+b2a)2−(b2−4ac√2a)2=0

Now once again if you notice (m+n)(m-n)= m^2-n^2, in relating to above , the equation can be rewritten as

(x+b2a+b2−4ac√2a)(x+b2a−b2−4ac√2a)=0

Now when will the above equation be zero? If either of the terms in brackets are zero.

Therefore,

(x+b2a+b2−4ac√2a)=0OR(x+b2a−b2−4ac√2a)=0=>x=−b2a−b2−4ac√2aORx=−b2a+b2−4ac√2a

Now identify the a,b,c in your equation substitute in solution you just found and you will get -1+√3 and -1-√3