Question

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? mc010-1.jpg mc010-2.jpg mc010-3.jpg mc010-4.jpg

Answer 1

Answer: Using the quadratic formula to solve the quadratic equation 5x=6x^2-3, the values of x are:

x1=[5-sqrt(97)]/12

x2=[5+sqrt(97)]/12

Solution:

5x=6x^2-3

To apply the quadratic formula, our quadratic equation must have the form:

ax^2+bx+c=0, with all the terms on one side of the equation and equaling the equation to zero.

5x=6x^2-3

Subtracting 5x both sides of the equation:

5x-5x=6x^2-3-5x

0=6x^2-5x-3

6x^2-5x-3=0

Comparing with ax^2+bx+c=0→a=6, b=-5, c=-3

Using the quadratic formula:

x=[-b+-sqrt(b^2-4ac)] / (2a)

Repacing the values of a, b, and c:

x=[-(-5)+-sqrt( (-5)^2-4(6)(-3) )] / (2(6))

x=[5+-sqrt(25+72)]/12

x=[5+-sqrt(97)]/12

x1=[5-sqrt(97)]/12

x2=[5+sqrt(97)]/12

Answer 2

Answer:

$x=\frac{5{\pm}\sqrt{97}}{12}$.

Step-by-step explanation:

The given equation is:

$5x=6x^2-3$

Upon rearranging the terms of the above equation, we get

$6x^2-5x-3=0$

Using the Quadratic formula that is $x=\frac{-b{\pm}\sqrt {b^2-4ac}}{2a}$, we have

$x=\frac{-(-5){\pm}\sqrt{(-5)^2-4(6)(-3)}}{2(6)}$

$x=\frac{5{\pm}\sqrt{25+72}}{12}$

$x=\frac{5{\pm}\sqrt{97}}{12}$

therefore, the value of x of the given equation is:

$x=\frac{5{\pm}\sqrt{97}}{12}$.