Question

the roots of x^2 + ax + b = 0 is K and L . the roots of x^2 + px + q = 0 is K and N. the roots of x^2 + bx+ a = 0 is T and S . the roots of x^2 + qx + p = 0 is T and R ... if a , b, p , q are real constants then prove
a+ b = p + q

Answer 1

@ganeshie8 and @phi pls help

Answer 2

Factor all the equations in terms of the roots:
\[
x^2+ax+b=(x-K)(x-L)=0 \\
x^2+px+q=(x-K)(x-N)=0 \\
x^2+bx+a=(x-T)(x-S)=0 \\
x^2+qx+p=(x-T)(x-R)=0
\]
Now subtract the second from the first, and the last from the second to last, and factor the expressions:
\[
(x-K)((x-L)-(x-N))=(x-K)(N-L)=0 \\
(x-T)((x-S)-(x-R))=(x-T)(R-S)=0
\]
In both expressions, one factor must be zero, and it can't be the one containing \(x\) since \(x\) is variable. Thus
\[N-L=0\Leftrightarrow N=L \\
R-S=0\Leftrightarrow R=S\]
Putting this into our original factorization above, we see that the first two and the last two equations are respectively equal. Specifically, the coefficients must be equal, so \(a=p\) and \(b=q\). From this it immediately follows that
\[a+b=p+q\]

Answer 3

I think we can do this without regard to the roots:
x^2 + ax + b = 0
x^2 + bx+ a = 0
add these equations to get
2x^2 +(a+b)x +(a+b) = 0 true for all x

similarly
x^2 + px + q = 0
x^2 + qx + p = 0

2x^2 +(p+q)x + (p+q) = 0 true for all x

equating zero to zero:
2x^2 +(a+b)x +(a+b) =2x^2 +(p+q)x + (p+q)

(a+b)x +(a+b) =(p+q)x + (p+q)

the left side (a line) coincides the line on the right side (for all x) only if the slopes and intercepts are equal. i.e. only when a+b= p+q