Question

written work
do these problems on some clean paper. label each page of your work with your name, your class, the date and the book number. also number each problem. keep this written work inside your book, and turn it in with your book when you are finished. please do a neat job.

1. explain why 1.4142135 ≈ √2.
2. copy each expression which stands for a rational number and tell what rational number it equals.

√5 √25 √20 √20√5 √5√5 √20/√5 √25/√5 √25/√4
-√5 -√4 -√4 -√1 √0 √1 √10 √100 √1000 √10000

1. simplify each expression.

√50 √3 √(1/7) √(5/9) √(5/6) 2√12 3/√2 √6x² √49x √24 + √600

1. approximate to the nearest hundredth.

√42 3√5 4 - √10 √6 + √2

1. solve each equation. if the solutions are not rational numbers, give them in simplest radical form.

x² = 26 (x - 2)² = 19 3x² - 1 = 11
x² + 2x - 8 = 0 x² + 8x - 2 = 0 2x² - 5x + 3 = 0

1. solve the equation x² - 8x = 0 in three different ways.
2. find the lengths of the sides of each triangle or rectangle to the nearest hundredth.
3. try to solve x² - 4x + 7 = 0 by using the quadratic formula. what happens?
4. graph the quadratic function f(x)=x² - 4x + 7, using -1, 0, 1, 2, 3, 4 and 5 as the x - values in your table. what does the result have to do with the result in problem 8?

Explanation:

1. Recall the nature of irrational numbers and approximations.
2. Use the definition of square - roots and properties of square - roots for rational numbers.
3. Apply the rules of simplifying square - root expressions such as $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}(a\geq0,b\geq0)$ and rationalizing the denominator.
4. Use a calculator to approximate the square - root values and perform arithmetic operations.
5. Solve quadratic equations by factoring, completing the square, and the quadratic formula.
6. Solve the quadratic equation using different methods based on factoring, completing the square, and the quadratic formula.
7. Use the Pythagorean theorem for right - triangles and area formulas for rectangles and then solve the resulting equations.
8. Apply the quadratic formula and analyze the discriminant.
9. Relate the graph of a quadratic function to the solutions of the corresponding quadratic equation by considering the discriminant and the vertex of the parabola.

Answer:

1. The decimal 1.4142135 is an approximation of $\sqrt{2}$. The square - root of 2 is an irrational number, and its decimal expansion is non - repeating and non - terminating. When we calculate $\sqrt{2}$ using a calculator or other methods, one of the early approximations is 1.4142135.

2.

• $\sqrt{25}=5$ (since $5\times5 = 25$)
• $\sqrt{0}=0$ (since $0\times0 = 0$)
• $\sqrt{1}=1$ (since $1\times1 = 1$)
• $\sqrt{100}=10$ (since $10\times10 = 100$)
• $\sqrt{10000}=100$ (since $100\times100 = 10000$)
• $\frac{\sqrt{25}}{\sqrt{5}}=\sqrt{\frac{25}{5}}=\sqrt{5}$ (using the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}},a\geq0,b > 0$)
• $\frac{\sqrt{25}}{\sqrt{4}}=\frac{5}{2}$ (since $\sqrt{25}=5$ and $\sqrt{4}=2$)

3.

• $\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}$
• $\sqrt{\frac{1}{7}}=\frac{\sqrt{1}}{\sqrt{7}}=\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}$
• $\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{\sqrt{9}}=\frac{\sqrt{5}}{3}$
• $2\sqrt{12}=2\sqrt{4\times3}=2\times2\sqrt{3}=4\sqrt{3}$
• $\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$ (by rationalizing the denominator: $\frac{3}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$)
• $\sqrt{6x^{2}}=\vert x\vert\sqrt{6}$
• $\sqrt{49x}=7\sqrt{x},x\geq0$
• $\sqrt{24}+\sqrt{600}=\sqrt{4\times6}+\sqrt{100\times6}=2\sqrt{6}+10\sqrt{6}=12\sqrt{6}$

4.

• $\sqrt{42}\approx6.48$
• $3\sqrt{5}\approx3\times2.24 = 6.72$
• $4-\sqrt{10}\approx4 - 3.16=0.84$
• $\sqrt{6}+\sqrt{2}\approx2.45+1.41 = 3.86$

5.

• For $x^{2}=26$, $x=\pm\sqrt{26}$
• For $(x - 2)^{2}=19$, $x-2=\pm\sqrt{19}$, so $x = 2\pm\sqrt{19}$
• For $3x^{2}-1 = 11$, $3x^{2}=12$, $x^{2}=4$, $x=\pm2$
• For $x^{2}+2x - 8=0$, factor as $(x + 4)(x - 2)=0$, so $x=-4$ or $x = 2$
• For $x^{2}+8x - 2=0$, using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b = 8$, $c=-2$, $x=\frac{-8\pm\sqrt{64+8}}{2}=\frac{-8\pm\sqrt{72}}{2}=\frac{-8\pm6\sqrt{2}}{2}=-4\pm3\sqrt{2}$
• For $2x^{2}-5x + 3=0$, factor as $(2x - 3)(x - 1)=0$, so $x = 1$ or $x=\frac{3}{2}$

6.

• Method 1: Factoring
• $x^{2}-8x=x(x - 8)=0$, so $x = 0$ or $x = 8$
• Method 2: Completing the square
• $x^{2}-8x=0$, $x^{2}-8x+16=16$, $(x - 4)^{2}=16$, $x-4=\pm4$, so $x=0$ or $x = 8$
• Method 3: Quadratic formula
• For $x^{2}-8x=0$, where $a = 1$, $b=-8$, $c = 0$. Then $x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4\times1\times0}}{2\times1}=\frac{8\pm\sqrt{64}}{2}=\frac{8\pm8}{2}$, so $x = 0$ or $x = 8$

7.

• For the right - triangle with legs 7 and 11, by the Pythagorean theorem $c=\sqrt{7^{2}+11^{2}}=\sqrt{49 + 121}=\sqrt{170}\approx13.04$
• For the rectangle with area $A = 252$ and sides $x$ and $2x$, $A=2x\cdot x=2x^{2}=252$, $x^{2}=126$, $x=\sqrt{126}\approx11.22$, $2x\approx22.43$
• For the rectangle with area $A = 60$ and sides $x$ and $x + 5$, $x(x + 5)=60$, $x^{2}+5x-60=0$. Using the quadratic formula $x=\frac{-5\pm\sqrt{25+240}}{2}=\frac{-5\pm\sqrt{265}}{2}$. We take the positive root $x=\frac{-5+\sqrt{265}}{2}\approx5.69$, $x + 5\approx10.69$

1. For the quadratic equation $x^{2}-4x + 7=0$, where $a = 1$, $b=-4$, $c = 7$. The discriminant $\Delta=b^{2}-4ac=(-4)^{2}-4\times1\times7=16 - 28=-12<0$. So there are no real solutions.
2. When we graph the quadratic function $y=x^{2}-4x + 7$, we can find the vertex. The $x$ - coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{-4}{2\times1}=2$. Then $y=2^{2}-4\times2 + 7=4-8 + 7 = 3$. The graph of $y=x^{2}-4x + 7$ is a parabola opening upwards ($a = 1>0$) with vertex $(2,3)$. Since the discriminant of the quadratic equation $x^{2}-4x + 7=0$ is negative, the graph of the function $y=x^{2}-4x + 7$ does not intersect the $x$ - axis, which is consistent with the fact that there are no real solutions for the equation $x^{2}-4x + 7=0$.