Virtual Math Tutor

Math problem #245 (Solution)

2012-02-17T18:49:00.000-04:00

Solution to Math problem #245:

Employing Cramer's rule, we note first that since

$$\Delta_x = \left|\begin{array}{cc} 12 & - 8 \\ 15 & - 6 \end{array}\right|\not= 0$$

the system does not have solutions only if

$$\Delta = \left|\begin{array}{cc} a & - 8 \\ 2 & - 6 \end{array}\right| = - 6a + 16 = 0,$$

that is when $a = 8/3$.

Cramer's rule

2012-02-14T18:45:00.000-04:00

Consider the following system of linear equations

$$\begin{cases}a_1x + b_1y = c_1, \\ a_2x + b_2y = c_2,\end{cases}\quad\quad (1)$$

where $a_1^2 + b_1^2 \not=0$, $a_2^2 + b_2^2 \not=0$. Next, let us define the following three determinants

$$\Delta = \left|\begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}\right| = a_1b_2 - a_2b_1;$$

$$\Delta_x = \left|\begin{array}{cc} c_1 & b_1 \\ c_2 & b_2 \end{array}\right| = c_1b_2 - c_2b_1;$$

$$\Delta_y = \left|\begin{array}{cc} a_1 & c_1 \\ a_2 & c_2 \end{array}\right| = a_1c_2 - a_2c_1.$$

Then the system (1) has a unique solution if and only if the determinant $\Delta \not=0$. In this case the solution is given by the formulas

$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta},$$

which are called Cramer's formulas, or Cramer's rule. If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta \not=0$ is equivalent to

$$\frac{a_1}{a_2} \not= \frac{b_1}{b_2}.$$

Conversely, the system (1) has no solutions if and only if $\Delta = 0$ and at least one of the determinants $\Delta_x$ and $\Delta_y$ is not equal to zero.

If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta = 0$, $\Delta_x \not=0$ (or, $\Delta = 0$, $\Delta_y \not=0$) is equivalent to the following condition

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \not=\frac{c_1}{c_2}.$$

The system (1) has infinitely many solutions if and only if $\Delta = \Delta_x = \Delta_y = 0.$ If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta = \Delta_x = \Delta_y = 0$ is equivalent to the following condition

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} =\frac{c_1}{c_2}.$$

If the conditions $a_1^2 + b_1^2 \not= 0$, $a_2^2 + b_2^2 \not=0$ are not satisfied, then $\Delta = \Delta_x = \Delta_y = 0$ may not imply that the system (1) has infinitely many solutions. For example, all of the three determinants of the linear system

$$\begin{cases} 0\cdot x + 0\cdot y = 4, \\ 0\cdot x + 0\cdot y = 15\end{cases}$$

vanish, but the system has no solutions.

Happy Valentine's day!

2012-02-13T22:38:00.002-04:00

The heart of a mathematician

2012-02-13T22:18:00.002-04:00

Math problem #245

2012-02-10T17:02:00.002-04:00

Math problem: Find all values of the parameter $a$, for which the linear system

$$\begin{cases} ax - 8x = 12, \\ 2x - 6y = 15 \end{cases}$$

has no solutions.

Discussion and hints: Investigate the coefficient matrix of the system, employ Cramer's rule.

Solution

Math problem #244 (Solution)

2012-02-10T16:56:00.001-04:00

Solution to Math problem #244:

Since $(x^2 - x) - (2-x) = x^2 - 2$, the inequality in question can be rewritten as

$$|x^2 - x| > |x^2 - x| - |2 - x|.$$

In view of the basic algebraic inequality BI2, we conclude that the set of solution to the given inequality coincides with the set of solutions to the following inequality

$$(2-x)(x^2 - x) < 0,$$ or, $$(x-2)(x - \sqrt{2})(x + \sqrt{2}) > 0.$$

The last inequality holds true for $ - \sqrt{2} < x < \sqrt{2}$ and $x > 2$, which is also the solution to the original inequality.

Math problem #244

2012-02-09T11:21:00.000-04:00

Math problem: Solve the following inequality

$$|x^2 - x| < |2-x| + |x^2 - 2|.$$ Discussion and hints: Employ the Basic algebraic inequalities as appropriate.

Solution

Basic algebraic inequalities

2012-02-09T11:15:00.004-04:00

BI1
The inequality $|a+b|< |a|+|b|$ holds true if and only if $|ab|<0,$ where $a, b \in \mathbb{R}$.

BI2
The inequality $|a-b| > |a| - |b|$ holds true if and only if $(a-b)b < 0,$ where $a, b \in \mathbb{R}$.

BI3
The inequality $|a+b| \le |a| + |b|$ holds true for any $a, b \in \mathbb{R}$.

Math problem #243 (Solution)

2012-02-08T09:43:00.000-04:00

Solution to Math problem #243:

Introduce a new variable: $\sqrt{x} = t$, $t\ge 0$, with respect to which the original equation assumes the form of a quadratic equation:

$$(a-3)t^2 - 8t + a + 3 = 0. \quad\quad (1)$$

If $a-3 = 0$, then the quadratic equation reduces to a linear equation:

$$-8t = - 6,$$

or,

$$t = \frac{3}{4},$$

or,

$$x = \frac{9}{16}.$$

If $a - 3 \not = 0$, then we obtain a unique solution, provided the discriminant $D$ of the quadratic equation (1) vanishes: $D = 64-4(a+3)(a-3) = 0,$ which leads to two values of the parameter $a$: $a = \pm 5.$

Have new formula for cube root, says Agra mathematician

2012-02-07T11:29:00.004-04:00

Follow the link to read the article:

Have new formula for cube root, says Agra mathematician

Math problem #243

2012-02-07T10:49:00.002-04:00

Math problem: Determine for which values of the parameter $a$ the following equation

$$(a-3)x - 8\sqrt{x} + a+3 = 0$$

has a unique solution.

Discussion and hints: Introduce a new variable: $\sqrt{x} = t$.

Solution

Math problem #242 (Solution)

2012-02-06T15:54:00.001-04:00

Solution to Math problem #242:

Let $u$ and $v$ be the sums of all coefficients of the even and odd powers of $x$ respectively. Then, we have $f(1) = u + v$ and $f(-1) = u - v$, where $f(x) = (x^2 + 2x + 2)^{2012} + (x^2 - 3x - 3)^{2012}.$ This observation leads to the following system of algebraic equations:

$$\begin{cases}

u+ v = 5^{2012} + (-5)^{2012} = 2 5^{2012}, \\

u - v = 1^{2012} + 1^{2012} = 2.
\end{cases}
$$

Therefore $v = 5^{2012} - 1.$

World Maths Day 2012: Register Now

2012-02-06T15:38:00.002-04:00

For more information please visit Letsplaymath:

World Maths Day 2012: Register Now

Mathematicians: You must have at least 17 clues to solve Sudoku

2012-02-06T15:34:00.000-04:00

A recent mathematics study showed that you have to have at least 17 clues on a Sudoku grid in order for the puzzle to be solvable. You could make the game easier, by adding more clues. But if there are fewer than 17 clues, then the game becomes impossible to solve. In this video, mathematician James Grime explains how the researchers figured this out.

via Boingboing

Math problem #242

2012-02-02T13:01:00.004-04:00

Math problem: Consider the following polynomial

$$(x^2 + 2x + 2)^{2012} + (x^2 - 3x - 3)^{2012}.$$

Find the sum of all coefficients of the odd powers of $x$.

Discussions and hints: Let $f(x)$ be an arbitrary polynomial, that is

$$f(x) = a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n.$$

Then $f(1)$ is equal to the sum of all coefficients of the polynomial $f(x)$, while $f(-1)$ - to the difference between the sum of all coefficients of the even powers of $x$ and the sum of all coefficients of the odd powers of $x$.

Solution

The thing about 998,001 is...

2012-01-31T11:36:00.001-04:00

If you divide 1 by the number 998,001, you get a list of all the three digit numbers in order except 998. Like so:

via kottke.org

Math problem #241 (Solution)

2012-01-27T16:39:00.000-04:00

Solution to Math problem #241:

First, we determine the domain of the function on the left hand side, it is obviously $x \ge 0.$ Now, the equation reduces to the following systems

$$\begin{cases} x \ge 0, \\ \sqrt{x} = 0, \end{cases}$$

or,

$$\begin{cases} x \ge 0, \\ x^2 - 1 = 0. \end{cases}$$

Therefore the two admissible solutions are $x_1 = 0$ and $x_2 = 1$. The problem is now completely solved.

Math problem #241

2012-01-24T12:40:00.001-04:00

Math problem: Solve the equation

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

Discussion and hints: It is easy to make a mistake via the following argument

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

$$x=0, \quad \mbox{or} \quad x^2 - 1.$$

Therefore the answer is $x = -1, 0, 1$.

Try to avoid making this mistake by taking into the account the domain of the function on the left hand side of the equation.

Solution

Math problem #240 (Solution)

2012-01-23T16:56:00.003-04:00

Solution to Math problem #240:

Using the basic trigonometric formulas, we transform the original trigonometric equation as follows:

$$0 = 2(\sin x + \cos x) + \sin 2x + 1 = $$

$$2(\sin x + \cos x) + (2\sin x\cos x + \sin^2 + \cos^2 x) = $$

$$(\sin x + \cos x)^2 + 2(\sin x+\cos x), $$

or,

$$(\sin x + \cos x)^2 + 2(\sin x+\cos x)=0, $$

or,

$$(\sin x + \cos x)(\sin x + \cos x + 2) =0.$$

Hence, we arrive at the following trigonometric equations

$$ \sin x + \cos x = 0\quad\quad (1)$$

and

$$ \sin x + \cos x = -2. \quad\quad (2)$$

We rewrite the first trigonometric equation as follows:
$$\cos x = \cos\left(\frac{\pi}{2} + x\right),$$

since $-\sin x = \cos\left(\frac{\pi}{2} + x\right)$. In view of the trigonometric formula TF51, we have

$$x = \pm\left(\frac{\pi}{2} + x\right) + 2\pi n, \, n \in \mathbb{Z}.$$

which is equivalent to

$$0 = \frac{\pi}{2} + 2\pi n, \, n \in \mathbb{Z},$$

or

$$x = -\frac{\pi}{4} + \pi n, \, n \in \mathbb{Z}.$$

The first of the last two equations obviously has no solutions. Thus, we have the solution
$$x = -\frac{\pi}{4} + \pi n, \, n \in \mathbb{Z}.$$

Now, let us solve the trigonometric equation (2). Using the trigonometric formula TF36, we rewrite (2) as follows:

$$\sin\frac{\pi}{4}\sin x + \cos\frac{\pi}{4}\cos x = -\sqrt{2},$$

or,

$$\cos \left(\frac{\pi}{4} - x\right) = -\sqrt{2} < -1,$$

which does not have any solutions. Therefore we conclude that the solution to the original equation is

$$x = -\frac{\pi}{4} + \pi n, \, n \in \mathbb{Z}.$$

The problem is now completely solved.

How do the Japanese multiply?

2012-01-21T17:58:00.002-04:00

Math problem #240

2012-01-20T15:40:00.002-04:00

Math problem: Solve the following trigonometric equation

$$2(\sin x + \cos x) + \sin 2x + 1 = 0.$$

Discussion and hints: Use the basic trigonometric formulas as appropriate.

Solution

Math problem #239 (Solutions)

2012-01-13T15:05:00.001-04:00

Solution to Math problem #239:

We have to determine the variables $a$, $b$ and $c$ in the formula of the quadratic polynomial

$$f(x) = ax^2 + bx + c,$$

given the three values of $f(x)$. Substituting them into the formula above, we get the following system three linear equations with $a$, $b$ and $c$ as the unknowns:

$$\begin{cases} 100a - 10b + c = 9, \\ 36a - 6b + c = 7, \\4a + 2b + c = -9.
\end{cases} $$

Subtracting from the first equation the third, we get $96a - 12b = 18$, or $16a - 2b = 3$. Next, subtracting from the second equation the third, we get $32a - 8b = 16$, or $8a - 2b = 4$. Finally, subtracting from the equation $16a - 2b = 3$ the equation $8a - 2b = 4$, find that $a = - 1/8$. Substituting this value into the equation $8a - 2b = 4$, we obtain $b = - 5/2$ and substituting the values $a = - 1/8$ and $b = - 5/2$ into the third equation of the original linear system, we get $c = - 7/2$.

Therefore the quadratic polynomial is as follows:

$$f(x) = -\frac{1}{8}x^2 - \frac{5}{2}x - \frac{7}{2}.$$

Math problem #239

2012-01-10T11:35:00.002-04:00

Math problem: Consider the following function defined by a quadratic polynomial

$$f(x) = ax^2 + bx + c.$$

Determine the parameters $a$, $b$, $c$ if $f(-10) = 9$, $f(-6) = 7$, $f(2) = -9$.

Discussion and hints: Reduce the problem to the problem of solving a system of three linear equations in three variables.

Solution

"Pythagorean theorem and dimensional analysis" post is fixed

2012-01-10T00:06:00.000-04:00

I've finally fixed the old post Pythagorean theorem and dimensional analysis (all of the formulas got corrupted at some point).