Mhyra4
Answered

When a quadratic equation has no linear term (variable raised to the first power), the solution becomes simpler.

What are the two possible solutions to [tex]$5x^2 - 15 = 0$[/tex]?

Enter your two answers separated by a comma.



Answer :

To solve the quadratic equation [tex]\( 5x^2 - 15 = 0 \)[/tex], follow these detailed steps:

1. Isolate the term with [tex]\( x^2 \)[/tex]:
- Start by adding 15 to both sides of the equation to move the constant term to the right side:
[tex]\[ 5x^2 - 15 + 15 = 0 + 15 \][/tex]
Simplifying both sides, we get:
[tex]\[ 5x^2 = 15 \][/tex]

2. Solve for [tex]\( x^2 \)[/tex]:
- Divide both sides by 5 to isolate [tex]\( x^2 \)[/tex]:
[tex]\[ \frac{5x^2}{5} = \frac{15}{5} \][/tex]
Simplifying both sides, we get:
[tex]\[ x^2 = 3 \][/tex]

3. Solve for [tex]\( x \)[/tex]:
- Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm \sqrt{3} \][/tex]

4. Simplify the square root:
- The two possible solutions are the positive and negative square roots of 3:
[tex]\[ x_1 = \sqrt{3} \approx 1.7320508075688772 \][/tex]
[tex]\[ x_2 = -\sqrt{3} \approx -1.7320508075688772 \][/tex]

So, the two possible solutions to the equation [tex]\( 5x^2 - 15 = 0 \)[/tex] are:

[tex]\( 1.7320508075688772, -1.7320508075688772 \)[/tex]