Math junior: Quadratic equations

Syllabus
1

Algebraic Expansions

We call some equations identities when they hold for any values of the
variables involved. For example, let’s consider the expression:

(x + y)(x + y)

Expanding it using the distributive property:

= x^2 + xy + yx + y^2 = x^2 + 2xy + y^2

So we have the identity:

(x + y)^2 = x^2 + 2xy + y^2

Example 1. (3 + 5)^2 = 64, and indeed
3^2 + 2 \cdot 3 \cdot 5 + 5^2 = 9 + 30 + 25 = 64

Problem 1. *Expand

(a-b)^2

Example 2. Expand (x+5)^2. Using (a+b)^2 = a^2 + 2ab + b^2 with
a=x and b=5: (x+5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25.

Example 3. Expand (2y-3)^2. Using (a-b)^2 = a^2 - 2ab + b^2
with a=2y and b=3:
(2y-3)^2 = (2y)^2 - 2(2y)(3) + 3^2 = 4y^2 - 12y + 9.

Problem 2. Expand the following expressions:

  1. (k+7)^2

  2. (m-4)^2

  3. (3x+1)^2

  4. (5y-2)^2

  5. (\sqrt{p} + \sqrt{q})^2

Problem 3 ([Difference of squares]{style=“color: blue”}). *Expand

(x-y)(x+y)

Factorization

Notice how the last identity gives a really useful way to rewrite
x^2 - y^2 as a product of two terms. Factorization is the process of
breaking down an expression into a product of simpler expressions
(factors). It’s like reverse expansion.

Example 4. Factorize 3x + 6. We observe that both terms have a
common factor of 3. 3x + 6 = 3(x) + 3(2) = 3(x+2).

Example 5. Factorize x^2 - 5x. Both terms have a common factor of
x. x^2 - 5x = x(x) - 5(x) = x(x-5).

Example 6. Factorize 2x(x+1) + 5(x+1). Here, (x+1) is a common
factor. 2x(x+1) + 5(x+1) = (x+1)(2x+5).

Example 7. Factorize x^2 -y^2.

x^2 -y^2 = x^2 + (xy - xy) - y^2 = x(x+y) - y(x+y) = (x+y)(x-y)

Problem 4. Factorize the following expressions:

  1. 7y - 21

  2. 4a^2 + 8a

  3. ab + ac

  4. 3x(x-2) - 7(x-2)

  5. 5p^2q - 10pq^2

  6. x^2 + 2x + 1

  7. x^2 - 2x + 1

  8. x^2 + 6x + 5

More problems on difference of squares.

Example 8. Factorize x^2 - 49. This is in the form a^2 - b^2
where a=x and b=7. x^2 - 49 = x^2 - 7^2 = (x-7)(x+7).

Problem 5. Factorize 9y^2 - 16.

Problem 6. 2x^2 - 50

Problem 7. Factorize the following expressions:

  1. k^2 - 100

  2. 25m^2 - 1

  3. 49 - n^2

  4. 18p^2 - 50q^2

  5. (x+y)^2 - z^2

Squares and roots

Example 9. Solve x^2 = 9. To find x, we take the square root of
both sides. \sqrt{x^2} = \sqrt{9} |x| = 3 This means x can be 3
or x can be -3. So, x = \pm 3. It is crucial to remember the \pm
(plus or minus) when taking the square root of both sides of an
equation. This accounts for both positive and negative roots.

Notice that \sqrt{(-2)^2} = 2 and not -2? See:
\sqrt{(-2)^2} = \sqrt{4} = 2. The square root symbol \sqrt{} always
denotes the non-negative square root. Therefore, \sqrt{a^2} = |a|
(since a could be negative).

Example 10. Solve (x-3)^2 = 25. Take the square root of both
sides: \sqrt{(x-3)^2} = \sqrt{25} |x-3| = 5 This leads to two
possibilities: 1) x-3 = 5 \implies x = 8 2) x-3 = -5 \implies x = -2
Thus, the solutions are x=8 and x=-2.

Problem 8. Solve the following equations:

  1. y^2 = 49

  2. 2x^2 = 50

  3. (z+1)^2 = 16

  4. (2m-3)^2 = 81

  5. 3(n-5)^2 = 12

Completing the Square {#completing-the-square .unnumbered}

Sometimes, a quadratic expression is not a perfect square but can be
transformed into one by adding a constant term. This process is called
completing the square.

Example 11. Consider x^2 + 6x + 9. This can be written as
(x+3)(x+3) = (x+3)^2. Notice that 9 = (6/2)^2 = 3^2.

Example 12. What term should be added to x^2 + 8x to make it a
perfect square trinomial? A perfect square trinomial is of the form
(x+k)^2 = x^2 + 2kx + k^2. Comparing x^2 + 8x with x^2 + 2kx, we
have 2k = 8, so k=4. The term to be added is k^2 = 4^2 = 16. So,
x^2 + 8x + 16 = (x+4)^2. The general rule is to add
(\text{half of the coefficient of } x)^2.

Example 13 (Completing the Square). Solve x^2 + 6x - 7 = 0. Move
the constant term to the right side: x^2 + 6x = 7 Now, complete the
square on the left side. Half of the coefficient of x (which is 6) is
3, and 3^2 = 9. Add 9 to both sides: x^2 + 6x + 9 = 7 + 9
(x+3)^2 = 16 Now, take the square root of both sides:
\sqrt{(x+3)^2} = \sqrt{16} |x+3| = 4 So, x+3 = 4 or x+3 = -4.
x = 1 or x = -7. The solutions are x=1 and x=-7.

Problem 9. Solve the following equations by completing the square:

  1. x^2 + 4x - 5 = 0

  2. y^2 - 10y + 9 = 0

  3. z^2 + 2z - 24 = 0

  4. m^2 - 6m = 7

  5. p^2 + 12p = -20

Simplifying Radicals and Rationalizing Denominators {#simplifying-radicals-and-rationalizing-denominators .unnumbered}

Working with square roots is common in quadratic equations. It’s
important to simplify them. Recall that \sqrt{ab} = \sqrt{a}\sqrt{b}
for non-negative a,b.

Example 14. Simplify \sqrt{75}. Find the largest perfect square
factor of 75, which is 25.
\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}.

Example 15. Simplify \sqrt{12x^3}. Assume x \ge 0.
\sqrt{12x^3} = \sqrt{4 \cdot 3 \cdot x^2 \cdot x} = \sqrt{4}\sqrt{x^2}\sqrt{3x} = 2|x|\sqrt{3x}.
Since we assume x \ge 0, this is 2x\sqrt{3x}.

Example 16 (Rationalizing a Denominator). Rationalize the
denominator of \frac{3}{\sqrt{2}}. Multiply the numerator and
denominator by \sqrt{2}:
\frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.

Example 17. Rationalize the denominator of \frac{1}{3+\sqrt{5}}.
Multiply the numerator and denominator by the conjugate of the
denominator, which is 3-\sqrt{5}.
\frac{1}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{3-\sqrt{5}}{3^2 - (\sqrt{5})^2} = \frac{3-\sqrt{5}}{9-5} = \frac{3-\sqrt{5}}{4}.

Problem 10. Simplify the following expressions:

  1. \sqrt{108}

  2. \sqrt{45y^5} (assume y \ge 0)

  3. \frac{5}{\sqrt{7}}

  4. \frac{2}{4-\sqrt{3}}

  5. \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}}

Let’s proceed with some more advanced problems:

Problem 11. Show that
x - y = (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) for x,y \ge 0.

Example 18. Simplify \sqrt{3 + 2\sqrt{2}}.

We want to find a and b such that
\sqrt{3 + 2\sqrt{2}} = \sqrt{a} + \sqrt{b}. Squaring both sides:
3 + 2\sqrt{2} = (\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}. By
comparing coefficients: a+b = 3 ab = 2 We need two numbers that sum
to 3 and multiply to 2. These are 1 and 2. So,
\sqrt{3 + 2\sqrt{2}} = \sqrt{2} + \sqrt{1} = \sqrt{2} + 1. This
technique works for expressions of the form \sqrt{A \pm 2\sqrt{B}}. If
the number in front of the inner radical is not 2, we try to make it 2
by moving factors in or out of the square root.

Example 19. Solve the equation \sqrt{5x-1} + \sqrt{x-1} = 2.

First, isolate one radical: \sqrt{5x-1} = 2 - \sqrt{x-1} Square both
sides: (\sqrt{5x-1})^2 = (2 - \sqrt{x-1})^2
5x-1 = 4 - 4\sqrt{x-1} + (x-1) 5x-1 = 3 + x - 4\sqrt{x-1} Group
terms: 4x - 4 = -4\sqrt{x-1} Divide by 4: x - 1 = -\sqrt{x-1} For
this equation to hold, both sides must be non-negative. Since the right
side is -\sqrt{x-1}, for it to be non-negative, \sqrt{x-1} must be
0, which means x-1=0 \implies x=1. Let’s check x=1 in the original
equation:
\sqrt{5(1)-1} + \sqrt{1-1} = \sqrt{4} + \sqrt{0} = 2 + 0 = 2. So x=1
is a solution. What if x-1 \neq 0? Then x-1 = -\sqrt{x-1} cannot
hold because the left side is positive (if x>1) and the right side is
negative. Thus, x=1 is the only solution.

Example 20. Simplify \sqrt{1 + \sqrt{21 + 12\sqrt{3}}}.

*First, simplify the inner nested radical \sqrt{21 + 12\sqrt{3}}. We
need to get a 2 in front of the inner radical:
12\sqrt{3} = 2 \cdot 6\sqrt{3} = 2\sqrt{36 \cdot 3} = 2\sqrt{108}. So,
\sqrt{21 + 12\sqrt{3}} = \sqrt{21 + 2\sqrt{108}}. We need two numbers
that sum to 21 and multiply to 108. Factors of 108:

(1, 108), (2, 54), (3, 36), (4, 27), (6, 18), (9, 12).

9+12=21 and
9 \times 12 = 108. So,
\sqrt{21 + 2\sqrt{108}} = \sqrt{12} + \sqrt{9} = 2\sqrt{3} + 3. Now
substitute this back into the original expression:
\sqrt{1 + (2\sqrt{3} + 3)} = \sqrt{4 + 2\sqrt{3}}. Again, use the
formula \sqrt{A + 2\sqrt{B}}. We need two numbers that sum to 4 and
multiply to 3. These are 1 and 3. So,
\sqrt{4 + 2\sqrt{3}} = \sqrt{3} + \sqrt{1} = \sqrt{3} + 1. Thus, the
simplified expression is \sqrt{3} + 1.*

Example 21. Simplify \sqrt{4+\sqrt{7}} - \sqrt{4 - \sqrt{7}}. Let
X = \sqrt{4+\sqrt{7}} - \sqrt{4 - \sqrt{7}}. Since
\sqrt{4+\sqrt{7}} > \sqrt{4 - \sqrt{7}}, X will be positive. Square
X: X^2 = (\sqrt{4+\sqrt{7}} - \sqrt{4 - \sqrt{7}})^2
X^2 = (4+\sqrt{7}) - 2\sqrt{(4+\sqrt{7})(4-\sqrt{7})} + (4-\sqrt{7})
X^2 = 4+\sqrt{7} - 2\sqrt{16-7} + 4-\sqrt{7} X^2 = 8 - 2\sqrt{9}
X^2 = 8 - 2(3) X^2 = 8 - 6 X^2 = 2 Since X is positive,
X = \sqrt{2}.

Problem 12. Simplify \sqrt{7 - 2\sqrt{10}}.

Problem 13. Simplify \sqrt{10 + \sqrt{84}}.

Problem 14. Solve the equation x + \sqrt{x+1} = 5.

Problem 15. If a, b, c are distinct non-negative numbers such
that \sqrt{a} + \sqrt{b} + \sqrt{c} = \sqrt{a+b+c}, prove that at
least one of \sqrt{a}, \sqrt{b}, \sqrt{c} must be zero. This implies
no such a,b,c exist.

Problem 16. Simplify
\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \ldots + \frac{1}{\sqrt{99}+\sqrt{100}}.