Form Ax B, Where A, B Are Real Numbers. 0

EQUATIONS
1. Linear equation in one variable

A linear equation in one variable is an equation that can be written in the
form
ax=b, where a, b are real numbers. 𝑎 ≠ 0.
For the linear equation 𝑎𝑥 = 𝑏:

a. If 𝑎 ≠ 0, there is a UNIQUE solution;
Exercise 1. In which value of a , the equation (𝑎 − 3)𝑥 = 5 has a unique solution?
Solution: 𝑎 − 3 ≠ 0
𝑎≠3
𝑎 ∈ (−∞; 3) ∪ (3; +∞)
b. If 𝑎 = 0 and 𝑏 = 0, there are INFINTE number of solutions;
Exercise 2. If the equation 𝑎𝑥 + 9 = 𝑏 − 5𝑥, has infinite number of solutions, find
the product 𝑎𝑏.
Solution: 𝑎𝑥 + 5𝑥 = 𝑏 − 9
(𝑎 + 5)𝑥 = 𝑏 − 9
I. 𝑎+5 = 0 II. 𝑏 − 9 = 0
𝑎 = −5 𝑏=9
𝑎𝑏 = −5 × 9 = −45
c. If 𝑎 = 0 and 𝑏 ≠ 0, there is NO solution.

Exercise 3. In which value of a, the equation
5(𝑥 − 𝑎) = 𝑎𝑥 − 10 doesn’t have a solution?
Solution:
5𝑥 − 5𝑎 = 𝑎𝑥 − 10
5𝑥 − 𝑎𝑥 = 5𝑎 − 10
(5 − 𝑎)𝑥 = 5𝑎 − 10
I. 5−𝑎 =0 II. 5𝑎 − 10 ≠ 0
𝑎=5 𝑎≠2
1
1. Number of solutions for Quadratic equations.
The discriminant is the part of the quadratic formula under the square root.
The discriminant can be positive, zero, or negative, and this determines how many
solutions there are to the given quadratic equation.
a. A positive discriminant indicates that the quadratic has two distinct real
number solutions.
Exercise 4. How many solutions (roots) does the following quadratic equation
have? 2𝑥 2 − 5𝑥 + 3 = 0
Solution. The discriminant should be more than 0. Let’s check.
𝑫 = 𝒃𝟐 − 𝟒𝒂𝒄 = (−𝟓)𝟐 − 𝟒 × 𝟐 × 𝟑 = 𝟏 > 𝟎
So, there are 2 distinct roots of the equation above.
b. A discriminant of zero indicates that the quadratic has a repeated real
number solution.
Exercise 5. Find the value of k for the following quadratic equation, so that
it has two equal roots.
Solution: We know that quadratic equation has two equal roots only when
the value of discriminant is equal to zero.
Putting discriminant equal to zero, we get
2
c. A negative discriminant indicates that neither of the solutions are real
numbers.
Exercise 6. In which values of m,
x 2 − 2(m + 3)x + 36 = 0 has no real solution?
Solution. In order not to have real solution, discriminant should be less than 0.
In the equation above, a=1, b= -2(m+3), c=36.
2
𝐷 = 𝑏2 − 4𝑎𝑐 = (−2(m + 3)) − 4 × 1 × 36 < 0
4(𝑚 + 3)2 − 144 < 0
Divide both sides by 4, to simplify calculations
(𝑚 + 3)2 − 36 < 0
𝑚2 + 6𝑚 + 9 − 36 < 0
𝑚2 + 6𝑚 − 27 < 0
(𝑚 + 9)(𝑚 − 3) < 0
Intercepts are m=-9 and m=3.
We investigate the sign of

(𝑚 + 9)(𝑚 − 3) in each interval formed by
intercepts and take negative interval as our
answer. So, in the interval (-9; 3) the
quadratic equation above does not have real solution.
3
2. Vieta’s Theorem
If 𝑥1 and 𝑥2 are roots of the quadratic equation 𝑥 2 + 𝑝𝑥 + 𝑞 = 0.

(coefficient of 𝑥 2 should be 1, a=1)
𝑥 + 𝑥2 = −𝑝
{ 1
𝑥1 𝑥2 = 𝑞
𝑥 + 𝑥2 = 5
Example. 𝑥 2 − 5𝑥 + 6 = 0. { 1 . So,
𝑥1 𝑥2 = 6
𝑥1 = 2, 𝑥2 = 3
Exercise 7. If x1 and x2 are roots of

x2 2
x 2 − 4x − 2 = 0 . Find x2 −
x1 +x2
Solution. According to Vieta’s theorem ,

𝑥1 + 𝑥2 = 4
{
𝑥1 𝑥2 = −2
Then,
x2 2 x2 (x1 +x2 )−x2 2 x1 x 2 −2 1
x2 − = = = =−
x1 +x2 x1 +x2 x1 +x2 4 2
3. Rational Equations.
𝑃(𝑥)
Equations in the form = 0 ( where P(x) and Q(x) are polynomials)
𝑄(𝑥)
Are called RATIONAL equations.
To solve rational equation:
1. P(x)=0 should be solved and roots are found.
2. 𝑄(𝑥) ≠ 0 should be solved. If it is among roots of P(x)=0, we do not take
it as solution of rational equation.
Exercise 8.
Solution.
4
4. Irrational Equations
Any equation where the variable is inside a radical is called an irrational
equation. When solving an irrational equation, the key step will be removing the
radical. To do that, we have to isolate each radical in one of the sides of
the equation, and then we have to do the square of both sides.
Exercise 9. Solve √𝑥 − 5 =3
Solution:
2
(√𝑥 − 5) =32
𝑥−5 = 9
𝑥 = 14
5
5. Absolute value equations
Absolute value equations are equations where the variable is within
an absolute value operator, like |x-5|=9.
If |𝒙| = 𝒃, we have 3 cases
I. 𝑰𝒇 𝒃 < 𝟎, 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆𝒓𝒆 𝒊𝒔 𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
II. 𝑰𝒇 𝒃 = 𝟎, 𝒕𝒉𝒆𝒏 𝒙 = 𝟎
𝒙=𝒃
III. 𝑰𝒇 𝒃 > 𝟎 {
𝒙 = −𝒃
Exercise 10. Solve: |𝑥 − 3| = 2
Solution: 2>0, So, we have 2 cases
a) x-3=2 b) x-3=-2
x=5 x=1
Solutions are 5 and 1.

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EQUATIONS

1. Linear equation in one variable

For the linear equation 𝑎𝑥 = 𝑏:

c. If 𝑎 = 0 and 𝑏 ≠ 0, there is NO solution.

Putting discriminant equal to zero, we get

We investigate the sign of

If 𝑥1 and 𝑥2 are roots of the quadratic equation 𝑥 2 + 𝑝𝑥 + 𝑞 = 0.

Exercise 7. If x1 and x2 are roots of

Solution. According to Vieta’s theorem ,

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