x2+7x+12=0
x2+7x=-12
x+Sqaure Root of 7x= 2(SQ Root Symbol)3
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x2+7x+12
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2
1
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)