Learn – Algebra foundations

Quadratic vs Linear expressions

Exercises

Problem set

Identify whether each of the following is a linear expression or a quadratic.

$-7x^2$
$-5x$
$5-9x^2$
$x$
$x^2$

Problem set

Identify whether each of the following is a linear expression or a quadratic.

$5^2x-7$
$7^2x+5x-9$
$3x(4x+1)$
$5x(3x-1)-2(3x-1)$
$(3x+1)(6x+5)$

Factoring

Exercises

Problem set

Write each of the following as a product of factors.

$3x(2x-1)+2(2x-1)$
$x(9x-4)-3(9x-4)$
$x(7x-3)-3(7x-3)$
$3x(2x-1)-(2x-1)$
$5x(3x+4)-(3x+4)$

Problem set

Write each of the following as a product of factors.

$x^2+4x+3x+12$
$2x^2+4x+3x+6$
$2x^2-3x+4x-6$
$9x^2+3x+6x+2$
$6x^2-4x+3x-2$
$8x^2-8x+x-1$

Problem set

Write each of the following as a product of factors.

$6x^2+4x-9x-6$
$7x^2+14x-9x-18$
$4x^2+2x-6x-3$
$12x^2+4x-6x-2$
$9x^2+3x-3x-1$

Problem set

Write each of the following as a product of factors.

$8x^2-6x-12x+9$
$9x^2-18x-2x+4$
$12x^2-4x-9x+3$
$7x^2-7x-9x+9$
$6x^2-6x-x+1$

Problem set

Write each of the following as a product of factors.

$6x^2+7x+2$
$2x^2-9x+9$
$3x^2-10x+3$
$4x^2-4x+1$
$5x^2-8x-4$

Problem set

Write each of the following as a product of factors.

$6x^2-5x+1$
$2x^2+5x-3$
$x^2-5x-6$
$3x^2+5x-2$
$-2+8x-6x^2$

Solving equations

By factoring

Exercises

Problem set

In each of the following equations, solve for the variable.

$(x+3)(x-8) = 0$
$(2x-1)(3x+7) = 0$
$(1-3x)(-5-2x) = 0$
$(2-7x)(5-3x) = 0$
$3(7x+5)(6x-5) = 0$
$(1-2x)(x+3) = 6$
$(3x+1)(2-x) = 4$

Problem set

In each of the following, solve for the variable by factoring.

$4x^2-5x+1=0$
$9x^2+12x+4=0$
$7x^2+12x-4=0$
$6x^2-5x-1=0$
$3+4x-4x^2=0$

Problem set

In each of the following, solve for the variable by factoring.

$2x^2-x=1$
$3x^2+14x=5$
$7x^2=15x-2$
$4x^2=x+3$
$3x^2-5x-6=2x-2x^2$
$(x+1)^2=9x-5$

With variable in a whole square

Exercises

Problem set

In each of the following equations, solve for the variable.

$x^2=25$
$2x^2 = 32$
$5x^2 = 45$
$9x^2-36 = 0$
$4x^2-9 = 0$

Problem set

In each of the following equations, solve for the variable.

$(x-2)^2=25$
$(2x+3)^2=1$
$(2x-1)^2=9$
$4(x-1)^2=25$
$9(3x-1)^2=16$
$-4(2-7x)^2=-9$
$-9(3-5x)^2=-25$

Problem set

In each of the following equations, solve for the variable.

$4(x+2)^2-9=0$
$9(x+7)^2-1=0$
$-16(5x-1)^2+25=0$
$-25(2x+1)^2+9=-16$
$-4(2x+1)^2+25=9$
$-4(2x-3)^2+3=0$
$-9(3x-1)^2+2=0$

By completing the square

Exercises

Problem set

In each of the following quadratic equations, solve for the variable by completing the square.

$x^2+4x+3=0$
$x^2+14x+13=0$
$x^2-6x-7=0$
$x^2-10x-11=0$
$x^2-12x+20=0$
$x^2-9x+14=0$
$x^2+5x-6=0$
$x^2-7x-8=0$

Problem set

In each of the following quadratic equations, solve for the variable by completing the square.

$2x^2+x-6=0$
$4x^2-7x+3=0$
$3x^2+4x+1=0$
$5x^2-4x-1=0$
$7x^2-2x-5=0$
$21x^2+4x-1=0$

Problem set

In each of the following quadratic equations, solve for the variable by completing the square.

$x^2-4x-2=0$
$x^2+6x+3=0$
$3x^2-x-1=0$
$4x^2+x-2=0$

Problem set

In each of the following quadratic equations, solve for the variable by completing the square.

$-3x^2+2x+2=0$
$-2x^2-2x+\frac{1}{2}=0$
$\frac{x^2}{2}-2x-1 = 0$
$6x^2-1=0$
$-36x^2+2=0$

Using the formula

Exercises

Problem set

In each of the following quadratic equations, solve for the variable using the quadratic formula.

$2x^2+5x+3 = 0$
$6x^2+7x+2 = 0$
$5x^2-12x+4 = 0$
$x^2-3x+2 = 0$
$x^2-3x-2 = 0$
$\frac{x^2}{2}-2x-1 = 0$
$\frac{x^2}{2}-\sqrt{10}x-3 = 0$
$\sqrt{2}x^2-4x+\sqrt{8} = 0$
$\frac{x^2}{2}-\sqrt{8}x+2 = 0$

Problem set

For each of the following quadratic equations, how many real solutions exist?

$3x^2-x+1 = 0$
$9x^2+3x+\frac{1}{4} = 0$
$9x^2+3x-1 = 0$

Whole square and difference of squares

This section uses the following formulas.

$(a+b)^2 = a^2+2ab+b^2$
$(a-b)^2 = a^2-2ab+b^2$
$(a+b)(a-b) = a^2-b^2$

Exercises

Problem set

Use the formulas of this section to get rid of parentheses in each of the following.

$(x+2)^2$
$(x-5)^2$
$(3x+1)^2$
$(2x-1)^2$
$\left(x+\frac{1}{5}\right)^2$
$\left(x-\frac{1}{3}\right)^2$
$\left(x-\frac{1}{5}\right)^2$
$\left(2x+\frac{1}{2}\right)^2$
$(-2x+1)^2$
$(-3x+2)^2$

Problem set

Use the formulas of this section to get rid of parentheses in each of the following.

$(r^2+2)^2$
$(r^2-2)^2$
$(r^2-r)^2$
$(x^3-x)^2$

Problem set

Use the formulas of this section to write each of the following as a whole square.

$x^2+6x+9$
$x^2+x+\frac{1}{4}$
$x^2-14x+49$
$4x^2-4x+1$
$x^2-\frac{2x}{3}+\frac{1}{9}$
$4x^2+9-12x$
$9x^2+\frac{4}{9}-4x$

Problem set

Use the formulas of this section to simplify or write in an alternate form.

$(x+2)(x-2)$
$(2x+1)(2x-1)$
$9999\times 10001$
$x^2-9$
$9x^2-36$

Miscellaneous

Exercises

Problem set

The solutions of $9(ax+b)^2 - 4 = 0$ are $1$ and $2$ . Find the values of $a$ and $b$ .
A quadratic has integer coefficients. One root of the quadratic is $-2+\sqrt{5}$ . What is the quadratic’s other root?
A quadratic has rational coefficients. One root of the quadratic is $-\frac{5}{2}-2\sqrt{5}$ . What is the quadratic’s other root?
A quadratic has rational coefficients. One root of the quadratic is $\sqrt{5}\left(\sqrt{5}-1\right)$ . What is the quadratic?
Simplify: $a\left(x + \frac{b+\sqrt{b^2 - 4ac}}{2a}\right)\left(x+\frac{b-\sqrt{b^2-4ac}}{2a}\right)$ .
Given a quadratic with rational coefficients, if one of its zeros is irrational, could the other zero be rational?