quadratic equations

What is a Quadratic Equation?

Quadratic equations are the polynomial equations of degree 2 in one variable. The standard form of a quadratic equation in one variable is ax² + bx + c where a, b, c, ∈ R and a ≠ 0. The values of x satisfying the quadratic equation are the roots of the quadratic equation. The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.

In this lesson, we will cover different ways to solve quadratic equations.

Solving Quadratic Equation Using Factorization

Solve x² + 2x − 15 = 0

Step 1: Express the equation in the form ax² + bx + c. This equation is already in this form.

Step 2: Factorize ax² + bx + c.
x² + 2x − 15
x² + 5x − 3x −15 ⇒ x(x + 5) − 3(x + 5) 

Step 3: Put each factor = 0.
(x − 3)(x + 5) = 0

Step 4: Solve each resulting equation.
x − 3 = 0 ⇒ x = 3
x + 5 = 0 ⇒x = −5

Answer: x = 3, −5

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Solving Quadratic Equation Using Formula

Let the given quadratic equation be ax² + bx + c = 0, where a ≠ 0. Solving this equation we get the value of x as \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Thus the roots of the given equation are \(\frac{-b + \sqrt{b^2-4ac}}{2a}\), \(\frac{-b - \sqrt{b^2-4ac}}{2a}\)

Examining the nature of the roots
For quadratic equation ax² + bx + c = 0, a ≠ 0, b² − 4ac is called the discriminant. We can get information on the nature of the roots by finding the value of the discriminant.

If b2 − 4ac > 0	Roots are real and distinct. If b2 − 4ac is a perfect square, the roots are real, rational, and distinct. If b2 − 4ac is not a perfect square then roots are real, irrational; and distinct.
If b2 − 4ac  = 0	Roots are real and equal
If b2 − 4ac  < 0	Roots are imaginary

Example: 4x² + 6x + 10
here b = 6, a = 4, c = 10 therefore, \( {6^2-4\cdot4\cdot10} = (36 - 160) < 0 \)
The roots for this equation are unreal or imaginary.

Example: 4x² + 4x + 1
here b = 4, a = 4, c = 1 therefore, 4² − 4⋅4⋅1 = 0
The roots are real and equal.

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Reducing Equation to Quadratic Form

Many equations may not be given as polynomial of the second degree or of form ax²+ bx + c = 0. But they can be reduced to the quadratic equation using suitable algebraic transformation.

Example: Solve \(\sqrt{x+9} + 3= x\)

Move 3 to the right-hand side, therefore \(\sqrt{x+9} = x -3\)

Squaring both sides
\({(\sqrt{x+9})}^2 = {(x-3)}^2\) 
\(x+9 = x^2 - 6x + 9\\ x^2-7x = 0\\ x(x-7) = 0\\ x = 0, x = 7 \)

Since x = 0, doesn't satisfy this condition therefore x= 7 is the only root.

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Let us solve word problems involving quadratic equations.
Example: In an auditorium, the number of seats in each row is 8 less than the number of rows. How many seats are in each row if there are 609 seats in the auditorium?
Solution: Let the number of rows be x. So the number of seats in each row is x − 8. Therefore, x⋅(x − 8) = 609
x² −8x − 609 = 0 ⇒ x² − 29x + 21x − 609 = 0
x⋅(x−29) + 21⋅(x−29) = 0 ⇒ (x−29)⋅(x+21) = 0
as x cannot be negative therefore x = 29.

Number of seats in each row = 29 − 8 = 21