If f(x)=5x^2-3 and g(x)=x^2-4x-8, find (f-g)(x)

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

4x^2 + 4x + 3.

Step-by-step explanation:

(f - g)(x)

= f(x) - g(x)

= 5x^2 - 3 - (x^2 - 4x - 6)

= 5x^2 - 3 - x^2 + 4x + 6

= 4x^2 + 4x + 3.

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Read 2 more answers

3x^2 - 4x = -2<br> Solve

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Answer:

$x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$

Step-by-step explanation:

$3 {x}^{2} - 4x = - 2 \\ 3 {x}^{2} - 4x + 2 = 0 \\ equating \: it \: with \\ a {x}^{2} + bx + c = 0 \\ a = 3 \: \: b = - 4 \: \: c = 2 \\ {b}^{2} - 4ac \\ = {( - 4)}^{2} - 4 \times 3 \times 2 \\ = 16 - 24 \\ = - 8 \\ \ {b}^{2} - 4ac < 0 \\ \therefore \: given \: quadratic \: equation \: have \: \\ imaginary \: solutios. \\ \\ x = \frac{ - b \pm \sqrt{{b}^{2} - 4ac } }{2a} \\ = \frac{ - ( - 4) \pm \sqrt{ - 8} }{2 \times 3} \\ = \frac{ 4 \pm 2\sqrt{2} \: i }{2 \times 3} \\ = \frac{ 2 \pm \sqrt{2} \: i }{3} \\ \therefore \: x = \frac{ 2 - \sqrt{2} \: i }{3} \: or \: x = \frac{ 2 + \sqrt{2} \: i }{3} \: \\ \\ x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$