SOLVING QUADRATIC EQUATIONS

To solve a quadratic equation which is in the form

ax2 + bx + c = 0

we have three different ways.

(i) Factoring

(ii) Quadratic formula

(iii) Completing the square

Solving Quadratic Equations by Factoring

Problem 1 :

x² - 4x + 3 = 0

Solution :

x² - 4x + 3 = 0

Since the coefficient of x2 is 1, we take the constant alone and decompose it into two parts.

Factors of 3 :

The middle term alone is negative, so both factors will be negative.

x² - x - 3x + 3 = 0

x(x - 1) - 3(x - 1) = 0

(x - 1) (x - 3) = 0

x = 1 and x = 3

So, the solution is {1, 3}.

Problem 2 :

2x² - x = 12 + x

Solution : 

2x² - x = 12 + x

The given question is not in the standard form, so subtract 12 and x to get rid of it.

2x² - x - x - 12 = 0

2x² - 2x - 12 = 0

2(x² - x - 6) = 0

2(x² + 2x - 3x - 6) = 0

2(x + 2) (x - 3) = 0

(x + 2) (x - 3) = 0

x = -2 and x = 3

So, the solution is {-2, 3}.

Solving Quadratic Equation by Formula 

Problem 3 :

Solve

2x2 + x - 4 = 0

Solution :

x =-b ±b2-4ac2aa=2, b=1 and c=-4x =-1 ±12-4(2)(-4)2(2)x =-1 ±1+324x =-1 ±334

Problem 4 :

Using completing the square method, solve the equation.

x2 − 6x − 7 = 0

Solution :

x2 − 2⋅x⋅3 + 32 - 32 - 7 = 0

Write the coefficient of x as a multiple of 2.

(x - 3)2 - 32 - 7 = 0

(x - 3)2 - 9 - 7 = 0

(x - 3)2 - 16 = 0

Add 16 on both sides.

(x - 3)2 = 16

Take square roots on both sides, we get

x - 3 = √16

x - 3 =  ±4

x - 3 = 4 and x - 3 = -4

x = 4 + 3 and x = - 4 + 3

x = 7 and x = -1

Solving Quadratic Equations in Different Forms

Problem 5 :

3(x + 7)2 + 17 = 56

Solution :

The given question is not in the standard form.

3(x + 7)2 + 17 = 56

Subtract 17 on both sides.

3(x + 7)2 + 17 - 17 = 56 - 17

3(x + 7)2 = 39

Divide by 3 on both sides.

(x + 7)2 = 39/3

(x + 7)2 = 13

(x + 7) = 13

x + 7 = ±√13

x = ±√13 - 7

Problem 6 :

2x - 5√x + 2 = 0

Solution :

It is trinomial, even though it doesn't have square, by considering the middle term (√x). Let us try to write the other terms also in terms of √x.

2(√x)2 - 5√x + 2 = 0

Let √x = t

2t2 - 5t + 2 = 0

2t2 - 4t - 1t + 2 = 0

2t(t - 2) - 1(t - 2) = 0

(2t - 1) (t - 2) = 0

Equating each factor to 0, we get

t = 1/2 and t = 2

When t = 1/2 ==>  √x = 1/2 ==> x = 1/4

When t = 2 ==>  √x = 2 ==> x = 4

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