Describe and correct the error in finding the different of the polynomials ..

Mathematics

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

Am not seeing any error

$(4 {x}^{2} - x + 3) - (3 {x}^{2} - 5x - 6) \\ = (4 - 3) {x}^{2} - (5 + 1)x + (3 - 6) \\ = {x}^{2} - 6x - 3$

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$\text{Domain:}\\\\x-4\neq0\ \wedge\ x-2\neq0\ \wedge\ x^2-6x+8\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x^2-4x-2x+8\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x(x-4)-2(x-4)\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ (x-4)(x-2)\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x-4\neq0\ \wedge\ x-2\neq0\\\\\boxed{D:x\neq4\ \wedge\ x\neq2}$

$\dfrac{1}{x-4}+\dfrac{x}{x-2}=\dfrac{2}{x^2-6x+8}\\\\\dfrac{1(x-2)}{(x-4)(x-2)}+\dfrac{x(x-4)}{(x-4)(x-2)}=\dfrac{2}{(x-4)(x-2)}\\\\\dfrac{x-2+x(x-4)}{(x-4)(x-2)}=\dfrac{2}{(x-4)(x-2)}\iff x-2+x(x-4)=2\\\\x-2+(x)(x)+(x)(-4)=2\\\\x-2+x^2-4x=2\qquad\text{subtract 2 from both sides}\\\\x^2-3x-4=0\\\\x^2-4x+x-4=0\\\\x(x-4)+1(x-4)=0\\\\(x-4)(x+1)=0\iff x-4=0\ \wedge\ x+1=0\\\\x-4=0\qquad\text{add 4 to both sides}\\x=4\notin D\\\\x+1=0\qquad\text{subtract 1 from both sides}\\x=-1\in D$