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Question

Question: The solution of equation \(x^{2} + px + q = 0\)+\(x^{2} + qx + p = 0\)+\(p + q + 1 =\) is....

The solution of equation

x2+px+q=0x^{2} + px + q = 0+x2+qx+p=0x^{2} + qx + p = 0+p+q+1=p + q + 1 = is.

A

x2+ax+10=0x^{2} + ax + 10 = 0

B

x2+bx10=0x^{2} + bx - 10 = 0

C

a2b2a^{2} - b^{2}

D

x211x+ax^{2} - 11x + a

Answer

x2+ax+10=0x^{2} + ax + 10 = 0

Explanation

Solution

We have x2+ax+b=0x^{2} + ax + b = 0= x2+bx+a=0x^{2} + bx + a = 0

(a+b)(a + b) =x(3/4)(log2x)2+(log2x)5/4=2x^{(3/4)(\log_{2}x)^{2} + (\log_{2}x) - 5/4} = \sqrt{2}

a>0,b>0,c>0ax2+bx+c=0a > 0,b > 0,c > 0ax^{2} + bx + c = 0

kk 2x2kx+x+8=02x^{2} - kx + x + 8 = 0

2x2+3x+1=0l,m,n2x^{2} + 3x + 1 = 0l,m,n.