Solution to some quadratic equation for some educational level, the guide to follow when solving some exercise which you may come across… ## Solve the following quadratic equation

• 3x^2 + 5x -2=0
• 4x^2 + 14x =0
• 4x^2=9
• (2x-1) ^2 =1

SOLUTION

1. 3x^2 + 5x – 2 = 0

The product of the first and last term

3x^2 * -2 = -6x^2

The factors of the figure above which should give 5x when added or subtracted

6x + -x = +5x

Substitute the above factors into the equation

3x^2 + 6x – x – 2 = 0

3x(x+2) -1(x+2) = 0

(3x-1) (x+2) =0

Then so for the value of X with the brackets

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

X=1/3

Or

(X-2) =0 => x= -2

The values of x are 1/3 or -2

SOLUTION

2. 4x^2 + 14x =0

Their common highest factor is 2x

You factorize the above at the left hand side

4x^2 + 14x=0

2x(2x + 7)=0

2x=0

X=0/2

X=0

Or

2x + 7 =0

Subtract 7 from both side

2x +7-7 = 0-7

2x= -7

Making x subject formula

X= -7/2

X=-3 ½

The values of x are 0 or -3 ½

SOLUTION

3. 4x^2=9

Divide both sides with the coefficient of x^2

4x^2 /4 = 9/4

X^2= 9/4

Remove square by introducing square roots

√X^2 = ±√9/4

X =± 3/2

The values of x are +3/2 or -3/2

SOLUTION

4. (2x-1) ^2 =1

Factorize the left hand side (LHS)

(2x-1) (2x-1) = 1

Expand to remove brackets

4x^2 – 2x -2x +1 = 1

Collect like terms

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

4x^2 – 4x =0

Their highest common factor is 4x

4x^2 -4x =0

4x(x-1) =0

4x =0 => x= 0/4 =0

Or

X -1 =0

Add 1 to both side of the equation as shown below

X -1+1 =0+1

X =1

The values of x are 0 or 1 