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2 years ago

What are the solutions to 0=x^2-x+1?

Mathematics

1 answer:

kumpel [21]2 years ago

4 0

Answer: No solutions

Step-by-step explanation:

There are no solutions to 0 = x² - x + 1.

[Method One]

We can solve this by rewriting the equation, but it is quicker to graph it when there are no solutions. See attached.

[Method Two]

I said we can solve this by solving an equation, here it is. We will use the quadratic formula since the equation given is already in the proper form.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

$\displaystyle x=\frac{1\pm\sqrt{(-1)^2-4(1)(1)} }{2(1)}$

$\displaystyle x=\frac{1\pm\sqrt{-3} }{2}$

-> Negative numbers don't have real square roots, so there is no solution

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Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]

Zina [86]

Answer: Absolute minimum: f(-1) = -2 $\sqrt{6}$

Absolute maximum: f( $\sqrt{12.5}$ ) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = $t\sqrt{25-t^{2}}$

f'(t) = $1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)$

f'(t) = $\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }$

f'(t) = $\frac{25-2t^{2}}{\sqrt{25-t^{2}} }$ = 0

For this function to be zero, only denominator must be zero:

$25-2t^{2} = 0$

t = ± $\sqrt{2.5}$

$\sqrt{25-t^{2}}$ ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = $-\sqrt{2.5}$ and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = $t\sqrt{25-t^{2}}$

f(-1) = $-1\sqrt{25-(-1)^{2}}$

f(-1) = $-\sqrt{24}$

f(-1) = $-2\sqrt{6}$

f( $\sqrt{12.5}$ ) = $\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}$

f( $\sqrt{12.5}$ ) = 12.5

f(5) = $5\sqrt{25-5^{2}}$

f(5) = 0

Therefore, absolute maximum is f( $\sqrt{12.5}$ ) = 12.5 and absolute minimum is

f(-1) = $-2\sqrt{6}$ .

8 0

3 years ago

What are the solutions of the equation x4 - 9x2 + 8 = 0? Use u substitution to solve.

Sedbober [7]

Answer:

$\large\boxed{B.\ x=\pm1\ and\ x=\pm2\sqrt2}$

Step-by-step explanation:

$x^4-9x^2+8=0\\\\x^{2\cdot2}-9x^2+8=0\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(x^2)^2-9x^2+8=0\\\\\text{substitute}\ x^2=t\geq0\\\\t^2-9t+8=0\\\\t^2-t-8t+8=0\\\\t(t-1)-8(t-1)=0\\\\(t-1)(t-8)+0\iff t-1=0\ \vee\ t-8=0\\\\t-1=0\qquad\text{add 1 to both sides}\\t=1\geq0\\\\t-8=0\qquad\text{add 8 to both sides}\\t=8\geq0$

$t=x^2\to x^2=1\ \vee\ x^2=8\\\\x^2=1\Rightarrow x=\pm\sqrt1\to x=\pm1\\\\x^2=8\Rightarrow x=\pm\sqrt8\to x=\pm\sqrt{4\cdot2}\to x=\pm\sqrt4\cdot\sqrt2\to x=\pm2\sqrt2$

4 0

3 years ago

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Point C is located 8 inches from Point A and 6 inches from Point B. Points A and B are 2 inches apart. which statement is true?

jek_recluse [69]

I think Point A and B are 2 inches apart or all of them is right

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4 years ago

EASY MATH QUESTION I will give brainiest to first correct answer!

GalinKa [24]

Answer:

11.98

Step-by-step explanation:

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3 years ago

School with 500 students has 30 math teachers. How many math teachers are needed for 750 students?

klemol [59]

Assuming a direct proportional variation.

$\bf \begin{array}{ccll}
students&teachers\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
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7 0

3 years ago

What are the solutions to 0=x^2-x+1?​

What are the solutions to 0=x^2-x+1?