4 years ago

What are the real roots of 2x^3+9x^2+4x-15=0

1 answer:

8 0

Well, luckily it is apparent that (x-1) is a root because when x=1 the equation is equal to zero. So we can divide the equation by that factor to find the other roots.

(2x^3+9x^2+4x-15)/(x-1)
2x^2 r 11x^2+4x-15
11x r 15x-15
15 r 0

(x-1)(2x^2+11x+15)=0

(x-1)(2x^2+6x+5x+15)=0

(x-1)(2x(x+3)+5(x+3))=0

(x-1)(2x+5)(x+3)=0

So the roots are x= -3, -2.5, 1

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Evaluate lim x approaches to 2 : (sqrt(6-x)-2)/(sqrt(3-x)-1)

solong [7]

Answer:

The answer is "0.5".

Step-by-step explanation:

Given:

$\to \lim_{x\to 2} \frac{(\sqrt{(6-x)}-2)}{(\sqrt{(3-x)}-1)}\\\\ \to \lim_{x\to 2} \frac{\frac{d(\sqrt{(6-x)}-2)}{dx}}{\frac{(\sqrt{(3-x)}-1)}{dx}}\\\\$

$\to \lim_{x\to 2} \frac{\frac{-1}{2\sqrt{(6-x)} }}{\frac{-1}{2\sqrt{(3-x)}}}\\\\$

$\to \lim_{x\to 2} \frac{\sqrt{3-x}}{\sqrt{6-x}}\\\\ \to \frac{\sqrt{3-2}}{\sqrt{6-2}}\\\\ \to \frac{\sqrt{1}} {\sqrt{4}}\\\\ \to \frac{1}{2}\\\\ \to 0.5$