How many solutions does 2x^2-3x+6=0 have?

Mathematics

2 answers:

Ipatiy [6.2K]3 years ago

8 0

Answer:

i beileve in this answer its a question of more than one salution, one salution and no solutions ................ so more than one salutions

Step-by-step explanation:

Katarina [22]3 years ago

4 0

Answer:

the solution of the equation will be two

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Gemiola [76]

Answer:

linear equations are hard but try L=16

Step-by-step explanation:

I'm prob wrong sorry

6 0

3 years ago

Find the poin on the graph 0f f(x)=1-x^2 that are closest to0(0,0)

zalisa [80]

Represent any point on the curve by (x, 1-x^2). The distance between (0, 0) and (x, 1-x^2) is

$\sqrt{(x-0)^2+(1-x^2-0)^2}=\sqrt{x^2+(1-x^2)^2}=\sqrt{x^2+1-2x^2+x^4}$

To make this easier, let's minimize the SQUARE of this quantity because when the square root is minimal, its square will be minimal.

So minimize $L=x^4-x^2+1$

Find the derivative of L and set it equal to zero.

$\frac{d}{dx}(L)=4x^3-2x \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0$

This gives you $x=0$ or $x^2=\frac{1}{2} \\ x=\pm\sqrt{2}/2$

You can use the Second Derivative Test to figure out which value(s) produce the MINIMUM distance.

$\frac{d^2}{dx}=12x^2-2$

When x = 0, the second derivative is negative, indicating a relative maximum. When $x=\pm\frac{\sqrt{2}}{2}$ , the second derivative is positive, indicating a relative MINIMUM.

The two points on the curve closest to the origin are $\left( \pm\frac{\sqrt{2}}{2},\frac{1}{2} \right)$