The area of the triangle is eqaula to the sqare show that x²-3x-2=0

Mathematics

1 answer:

agasfer [191]1 year ago

8 0

Answer:

Step-by-step explanation:

Area of square = area of triangle

$side*side = \dfrac{1}{2}b*h\\\\\\x*x=\dfrac{1}{2}*(x+1)(x+2)\\x^{2}=\dfrac{1}{2}[x*x +x*2+1*x+1*2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+2x+x+2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+3x+2]\\\\\\Multiply \ both \ sides \ by \ 2\\\\2x^{2}=x^{2}+3x+2\\\\2x^{2}-x^{2}-3x-2 = 0$

x² - 3x - 2 = 0

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The area of the triangle is eqaula to the sqare show that x²-3x-2=0​

The area of the triangle is eqaula to the sqare show that x²-3x-2=0