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3 years ago

Find ( f - g)(x). f (x) = -2x2 - 2x - 3 and g(x) = 3x2 - 4x - 4

Mathematics

1 answer:

Kruka [31]3 years ago

3 0

Answer:

$\large\boxed{(f-g)(x)=-5x^2+2x+1}$

Step-by-step explanation:

$(f-g)(x)=f(x)-g(x)\\\\\text{We have}\ f(x)=-2x^2-2x-3\ \text{and}\ g(x)=3x^2-4x-4.\\\\\text{Substitute:}\\\\(f-g)(x)=(-2x^2-2x-3)-(3x^2-4x-4)\\\\(f-g)(x)=-2x^2-2x-3-3x^2+4x+4\qquad\text{combine like terms}\\\\(f-g)(x)=(-2x^2-3x^2)+(-2x+4x)+(-3+4)\\\\(f-g)(x)=-5x^2+2x+1$

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frosja888 [35]

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Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

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$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

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$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

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$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

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