Chapter 5

Algebra 2 and Trigonometry · 322 exercises

Problem 1

Explain the relationship between the solutions of \(y>a x^{2}+b x+c\) and the solutions of \(0>a x^{2}+b x+c\)

5 step solution

Problem 1

Sharon said that if \(\mathrm{f}(x)\) is a polynomial function and \(\mathrm{f}(a)=0,\) then \(a\) is a root of the function. Do you agree with Sharon? Explain why or why not.

3 step solution

Problem 1

Emily said that when \(a\) and \(c\) are real numbers with the same sign and \(b=0\) , the roots of the equation \(a x^{2}+b x+c=0\) are pure imaginary. Do you agree with Emily? Justify your answer.

5 step solution

Problem 1

a. What is the discriminant of the equation \(x^{2}+(\sqrt{5}) x-1=0 ?\) b. Find the roots of the equation \(x^{2}+(\sqrt{5}) x-1=0\) c. Do the rules for the relationship between the discriminant and the roots of the equation given in this chapter apply to this equation? Explain why or why not.

6 step solution

Problem 1

The roots of a quadratic equation with rational coefficients are \(p \pm \sqrt{q} .\) Write the equation in standard form in terms of \(p\) and \(q .\)

5 step solution

Problem 1

Tim said that the binomial \(x^{2}+16\) can be written as \(x^{2}-16 i^{2}\) and factored over the set of complex numbers. Do you agree with Tim? Explain why or why not.

4 step solution

Problem 1

Pete said that \(\sqrt{-2} \times \sqrt{-8}=\sqrt{16}=4 .\) Do you agree with Pete? Explain why or why not.

4 step solution

Problem 1

Noah said that the solutions of Example \(1, \frac{2+\sqrt{6}}{2}\) and \(\frac{2-\sqrt{6}}{2},\) could have been written as \(\frac{2+\sqrt{6}}{2}=1+\sqrt{6}\) and \(\frac{-}{2}-\sqrt{6}=1-\sqrt{6} .\) Do you agree with Noah? Explain why or why not.

4 step solution

Problem 2

Explain why the equations \(y=x^{2}+2\) and \(y=-2\) have no common solution in the set of real numbers.

3 step solution

Problem 2

Noah said that if \(a, b,\) and \(c\) are rational numbers and \(b^{2}-4 a c<0\) , then the roots of the equation \(a x^{2}+b x+c=0\) are complex conjugates. Do you agree with Noah? Justify your answer.

4 step solution

Problem 2

Jordan said that if the roots of a polynomial function \(\mathrm{f}(x)\) are \(r_{1}, r_{2},\) and \(r_{3},\) then the roots of \(\mathrm{g}(x)=\mathrm{f}(x-a)\) are \(r_{1}+a, r_{2}+a,\) and \(r_{3}+a .\) Do you agree with Jordan? Explain why or why not.

4 step solution

Problem 2

Christina said that when \(a, b,\) and \(c\) are rational numbers and \(a\) and \(c\) have opposite signs, the quadratic equation \(a x^{2}+b x+c=0\) must have real roots. Do you agree with Christina? Explain why or why not.

5 step solution

Problem 2

Adrien said that if the roots of a quadratic equation are \(\frac{1}{2}\) and \(\frac{3}{4}\) , the equation is \(4 x^{2}-5 x+\frac{3}{2}=0 .\) Olivia said that the equation is \(8 x^{2}-10 x+3=0 .\) Who is correct? Justify your answer.

4 step solution

Problem 2

Joshua said that the product of a complex number and its conjugate is always a real number. Do you agree with Joshua? Explain why or why not.

5 step solution

Problem 2

Ethan said that the square of any pure imaginary number is a negative real number. Do you agree with Ethan? Justify your answer.

4 step solution

Problem 2

Rita said that when \(a, b,\) and \(c\) are real numbers, the roots of \(a x^{2}+b x+c=0\) are real numbers only when \(b^{2} \geq 4 a c .\) Do you agree with Rita? Explain why or why not.

3 step solution

Problem 2

Phillip said that the equation \(0=x^{2}-6 x+1\) can be solved by adding 8 to both sides of the equation. Do you agree with Phillip? Explain why or why not.

4 step solution

Problem 3

In \(3-14,\) use the quadratic formula to find the imaginary roots of each equation. $$ x^{2}-4 x+8=0 $$

5 step solution

Problem 3

In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ \mathrm{f}(x)=x^{3}+7 x^{2}+10 x $$

4 step solution

Problem 3

Without solving each equation, find the sum and product of the roots. \(x^{2}+x+1=0\)

4 step solution

Problem 3

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ (6+7 i)+(1+2 i) $$

5 step solution

Problem 3

In \(3-18,\) write each number in terms of \(i\) $$ \sqrt{-4} $$

5 step solution

Problem 3

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ x^{2}+5 x+4=0 $$

5 step solution

Problem 3

In \(3-8,\) complete the square of the quadratic expression. $$ x^{2}+6 x $$

5 step solution

Problem 4

In \(3-14,\) use the quadratic formula to find the imaginary roots of each equation. $$ x^{2}+6 x+10=0 $$

6 step solution

Problem 4

In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ f(x)=2 x^{3}+2 x^{2}-4 x $$

4 step solution

Problem 4

Without solving each equation, find the sum and product of the roots. \(x^{2}+4 x+5=0\)

3 step solution

Problem 4

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ (3-5 i)+(2+i) $$

4 step solution

Problem 4

In \(3-18,\) write each number in terms of \(i\) $$ \sqrt{-81} $$

3 step solution

Problem 4

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ x^{2}+6 x-7=0 $$

6 step solution

Problem 4

In \(3-8,\) complete the square of the quadratic expression. $$ x^{2}-8 x $$

5 step solution

Problem 5

In \(3-14,\) use the quadratic formula to find the imaginary roots of each equation. $$ x^{2}-4 x+13=0 $$

5 step solution

Problem 5

Without solving each equation, find the sum and product of the roots. \(2 x^{2}-3 x-2=0\)

5 step solution

Problem 5

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ (5-6 i)+(4+2 i) $$

5 step solution

Problem 5

In \(3-18,\) write each number in terms of \(i\) $$ \sqrt{-9} $$

3 step solution

Problem 5

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ x^{2}-3 x+1=0 $$

5 step solution

Problem 5

In \(3-8,\) complete the square of the quadratic expression. $$ x^{2}-2 x $$

5 step solution

Problem 6

In \(3-14,\) use the quadratic formula to find the imaginary roots of each equation. $$ 2 x^{2}+2 x+1=0 $$

6 step solution

Problem 6

Without solving each equation, find the sum and product of the roots. \(5 x^{2}+2 x-10=0\)

4 step solution

Problem 6

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ (-3+3 i)-(1+5 i) $$

4 step solution

Problem 6

In \(3-18,\) write each number in terms of \(i\) $$ -\sqrt{-36} $$

4 step solution

Problem 6

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ x^{2}-x-4=0 $$

4 step solution

Problem 6

In \(3-8,\) complete the square of the quadratic expression. $$ x^{2}-12 x $$

6 step solution

Problem 7

In \(3-14,\) use the quadratic formula to find the imaginary roots of each equation. $$ x^{2}+10 x+29=0 $$

6 step solution

Problem 7

In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ \mathrm{f}(x)=2 x^{3}-3 x^{2}-2 x+3 $$

6 step solution

Problem 7

Without solving each equation, find the sum and product of the roots. \(3 x^{2}-6 x+4=0\)

4 step solution

Problem 7

In \(3-17,\) find each sum or difference of the complex numbers in \(a+b i\) form. $$ (2-8 i)-(2+8 i) $$

5 step solution

Problem 7

In \(3-18,\) write each number in terms of \(i\) $$ -\sqrt{-121} $$

5 step solution

Problem 7

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ x^{2}+5 x-2=0 $$

5 step solution

Problem 7

In \(3-8,\) complete the square of the quadratic expression. $$ 2 x^{2}-4 x $$

4 step solution

Show/ page