The perimeter of a triangle is (6X2 -4X + 9) and two of its sides are (X2 -2X + 1) and (3X2 – 5X + 3). Find the third side of the triangle
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Let the third side be 
perimeter = sum of sides
x²-2x+1+3x²-5x+3+a = 6x²-4x +9
x²+3x²-6x²-2x-5x+4x+1+9+a = 0
-2x²-3x+10+a = 0
2x²+3x-10-a = 0
a = 2x²+3x-10
Therefore the third side is 2x²+3x-10
perimeter = sum of sides
x²-2x+1+3x²-5x+3+a = 6x²-4x +9
x²+3x²-6x²-2x-5x+4x+1+9+a = 0
-2x²-3x+10+a = 0
2x²+3x-10-a = 0
a = 2x²+3x-10
Therefore the third side is 2x²+3x-10
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Answer:
Let the third side be a.a.
perimeter = sum of sides
x²-2x+1+3x²-5x+3+a = 6x²-4x +9
x²+3x²-6x²-2x-5x+4x+1+9+a = 0
-2x²-3x+10+a = 0
2x²+3x-10-a = 0
a = 2x²+3x-10
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