# Find the roots of the equation 2x^2-3=5x using factors

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2x^2 - 3 = 5x

==> 2x^2 - 5x - 3 = 0

==> 2x^2 + x - 6x - 3 = 0

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

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

Roots of the equation are obtained by equating each of the two factor to 0.

When (2x + 1) = 0

x = -1/2

When (x - 3) = 0

x = 3

Answer:

Roots of the equation are -1/2 and 3

We'll re-write the equation having all terms to the left side:

2x^2 - 5x - 3 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25+24)]/4

x1 = (5+7)/4

x1 = 3

x2 = (5-7)/4

x2 = -1/2

We'll write the equation according as a product of linear factors:

2x^2 - 5x - 3 = 2(x - x1)(x - x2)

**2x^2 - 5x - 3 = 2(x - 3)(x + 1/2)**

We can also re-write the term 5x from the right side as the sum :2x + 3x.

Now, we'll re-write the entire equation:

2x^2-3 = 2x + 3x

We'll subtract 3x both sides and we'll add 3 both sides:

2x^2 - 3 + 3 - 3x = 2x + 3x + 3 - 3x

We'll combine and eliminate like terms:

2x^2 - 3x = 2x + 3

We'll factorize by x to the left side:

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

2x^2 -3= 5x

To solve for x.

subtract 5x:

2x^2-5x -3 = 0

Divide by 2:

x^2 -5x/2 -3/2 = 0

x^2-5x/2 = 3/2.

Add (5/4)^2 to both sides sothat LHS is a perfect square.

(x-5/4)^2 = (5/4)^2+3/2 = (25+24)/16 = 49/16

(x-5/4)^2 = (7/4)^2

x-5/2 = +7/4 Or x-5/2 = -7/4

x1 = 5/4+7/4 = 12/4 =3

x2 = (5/4 -7/4 = (10-7)/4 = -2/4 = -1/2

By factor metod:

2x^2-5x -3 = 0

2x^2-6x+x-3 = 0

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

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

x-3 = 0, Or 2x+1 = 0

x = 3 . Or x=-1/2