Question

What is the degree of 12x^2 + 5x^4 - 2?

Answer 1

3

Answer 2

3 isn't one of the options :( there's 2 or 4?

Answer 3

What is the higest power of x ?

Answer 4

oh sorry its 2 mybad

Answer 5

Solve for x over the real numbers:
5 x^4+12 x^2-2 = 0
Simplify 5 x^4+12 x^2-2 = 0 by making a substitution.
Substitute y = x^2:
5 y^2+12 y-2 = 0
Write the quadratic equation in standard form.
Divide both sides by 5:
y^2+(12 y)/5-2/5 = 0
Solve the quadratic equation by completing the square.
Add 2/5 to both sides:
y^2+(12 y)/5 = 2/5
Take one half of the coefficient of y and square it, then add it to both sides.
Add 36/25 to both sides:
y^2+(12 y)/5+36/25 = 46/25
Factor the left hand side.
Write the left hand side as a square:
(y+6/5)^2 = 46/25
Eliminate the exponent on the left hand side.
Take the square root of both sides:
y+6/5 = sqrt(46)/5 or y+6/5 = -sqrt(46)/5
Look at the first equation: Solve for y.
Subtract 6/5 from both sides:
y = sqrt(46)/5-6/5 or y+6/5 = -sqrt(46)/5
Perform back substitution on y = sqrt(46)/5-6/5.
Substitute back for y = x^2:
x^2 = sqrt(46)/5-6/5 or y+6/5 = -sqrt(46)/5
Eliminate the exponent on the left hand side.
Take the square root of both sides:
x = sqrt(sqrt(46)/5-6/5) or x = -sqrt(sqrt(46)/5-6/5) or y+6/5 = -sqrt(46)/5
Look at the third equation: Solve for y.
Subtract 6/5 from both sides:
x = sqrt(sqrt(46)/5-6/5) or x = -sqrt(sqrt(46)/5-6/5) or y = -6/5-sqrt(46)/5
Perform back substitution on y = -6/5-sqrt(46)/5.
Substitute back for y = x^2:
x = sqrt(sqrt(46)/5-6/5) or x = -sqrt(sqrt(46)/5-6/5) or x^2 = -6/5-sqrt(46)/5
Show the equation has no solution.
x^2 = -6/5-sqrt(46)/5 has no solution since for all x on the real line, x^2 >=0 and -6/5-sqrt(46)/5ɘ:
Answer: |
| x = sqrt(sqrt(46)/5-6/5) or x = -sqrt(sqrt(46)/5-6/5)

Answer 6

you do not have to solve the equation to find its degree

Answer 7

Degree = the highest power on the variable
in this case its 4