solve x2-2x-a(a+2)=0
Answers
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
Answer:
Step-by-step explanation:
x2-2x-a(a+2) = 0
-a(a+2) = 0-x2+2x
-a-2a = 0
-3a = 0
a = 0/-3
a = 0
Check:
Where a = 0,
(0×2)-(2×0)-0(0+2) = 0
0-0-0-0 = 0
Thenc proved
LHS = RHS
0 = 0