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3 years ago

Is 6/5 closer to 1/2

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

6 0

No, because 6/5 is a mixed number which means that it’ll have a whole number

Alecsey [184]3 years ago

5 0

$\frac{6}{5} =\quad 1.2$

$\frac{1}{2}= \quad 0.5$

$1.2 > 0.5$

$\frac{6}{5} > \frac{1}{2}$

<h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>

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Find m to cos²x-(m²-3)sinx+2m²-3=0 have root

vova2212 [387]

Answer:

$-\sqrt{2} \le m \le \sqrt{2}$ would ensure that at least one real root exists for this equation when solving for $x$ .

Step-by-step explanation:

Apply the Pythagorean identity $1 - \sin^{2}(x) = \cos^{2}(x)$ to replace the cosine this equation with sine:

$(1 - \sin^{2}(x)) - (m^2 - 3)\, \sin(x) + 2\, m^2 - 3 = 0$ .

Multiply both sides by $(-1)$ to obtain:

$-1 + \sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 3 = 0$ .

$\sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 2 = 0$ .

If $y = \sin(x)$ , then this equation would become a quadratic equation about $y$ :

$y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ .

$a = 1$ .
$b = m^{2} - 3$ .
$c = -2\, m^{2} + 2$ .

However, $-1 \le \sin(x) \le 1$ for all real $x$ .

Hence, the value of $y$ must be between $(-1)$ and $1$ (inclusive) for the original equation to have a real root when solving for $x$ .

Determinant of this quadratic equation about $y$ :

$\begin{aligned} & b^{2} - 4\, a\, c \\ =\; & (m^{2} - 3)^{2} - 4 \cdot (-2\, m^{2} + 2) \\ =\; & m^{4} - 6\, m^{2} + 9 - (-8\, m^{2} + 8) \\ =\; & m^{4} - 6\, m^{2} + 9 + 8\, m^{2} - 8 \\ =\; & m^{4} + 2\, m^{2} + 1 \\ =\; &(m^2 + 1)^{2} \end{aligned}$ .

Hence, when solving for $y$ , the roots of $y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ in terms of $m$ would be:

$\begin{aligned}y_1 &= \frac{-b + \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) + \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) + (m^{2} + 1)}{2} = 2\end{aligned}$ .

$\begin{aligned}y_2 &= \frac{-b - \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) - \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) - (m^{2} + 1)}{2} \\ &= \frac{-2\, m^{2} + 2}{2} = -m^{2} + 1\end{aligned}$ .

Since $y = \sin(x)$ , it is necessary that $-1 \le y \le 1$ for the original solution to have a real root when solved for $x$ .

The first solution, $y_1$ , does not meet the requirements. On the other hand, simplifying $-1 \le y_2 \le 1$ , $-1 \le -m^{2} + 1 \le 1$ gives:

$-2 \le -m^{2} \le 0$ .

$0 \le m^{2} \le 2$ .

$-\sqrt{2} \le m \le \sqrt{2}$ .

In other words, solving $y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ for $y$ would give a real root between $-1 \le y \le 1$ if and only if $-\sqrt{2} \le m \le \sqrt{2}$ .

On the other hand, given that $y = \sin(x)$ for the $x$ in the original equation, solving that equation for $x\!$ would give a real root if and only if $-1 \le y \le 1$ .

Therefore, the original equation with $x$ as the unknown has a real root if and only if $-\sqrt{2} \le m \le \sqrt{2}$ .

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