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3 years ago

Find x 5[x^2]+5[x]-x^2-x=2004

Mathematics

2 answers:

Vsevolod [243]3 years ago

5 0

Answer:

-22.89, 21.89 to the nearest hundredth.

Step-by-step explanation:

5[x^2]+5[x]-x^2-x=2004

5x^2 - x^2 + 5x - x - 2004 = 0

4x^2 + 4x - 2004 = 0

Divide through by 4:

x^2 + x - 501 = 0

Using the quadratic formula:

x = -1 +/- √(1^2 - 4*1*-501) / 2

x = -0.5 +/- 44.777/2

= -0.5 +/- 22.389

= -22.89, 21.89.

a_sh-v [17]3 years ago

4 0

The answer is either $x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\$ or $x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}$

To solve: Let's solve your equation step-by-step.

5x2+5x−x2−x=2004

Step 1: Simplify both sides of the equation.

4x2+4x=2004

Step 2: Subtract 2004 from both sides.

4x2+4x−2004=2004−2004

4x2+4x−2004=0

Step 3: Use quadratic formula with a=4, b=4, c=-2004.

x= −b±√b2−4ac /2a

x= −4±√32080 /8

$x=\frac{-1}{2}+\frac{1}{2}\sqrt{2005}\\$ or $x=\frac{-1}{2}+\frac{-1}{2}\sqrt{2005}$

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