Solve, what is the easiest way to solve this equation?
9x^4+4=13x^2 and
x^6-19x^3=216
Thanks, in advance..
I am solving the first one.....second one you can try yourself.
so 9x^4+4=13x^2
this can be written,
x^2*(9x^2-13)=-4
as x^2 is always +ve
so to be -ve,
(9x^2-13)<0
we see that
-(sqrt(13)/3)<x<(sqrt(13)/3)
it comes as -1.2<x<1.2
next we will see if there is some integer solution.
at x=1 and x=-1
the function has its value 0
so x=1 and x=-1 are solution to the equation.
next we will see if there is some maxima or minima in this func,(9x^4+4-13x^2)
after differentiating,
and taking it =0,
the func. has minima at +sqrt(13/18), -sqrt(13/18)
and at x=+1, x=-1 it has value 0,
so the another solutions are
x=1-2*sqrt(13/18) and x=-1+2*sqrt(13/18)
these are 4 solutions.
Ask me if you have any doubt.
There are two equations there.
In the first one, let y = x^2 and solve the quadratic.
9y^2 -13y +4 = 0
(y -1)(9y-4) = 0
y = 1 or 4/9
x = sqrt(y) = +/-1 and +/-2/3
In the second one, let y = x^3 and again solve the quadratic.
this does not help!
To solve the first equation, 9x^4 + 4 = 13x^2, we can use the quadratic equation by rearranging the terms:
9x^4 - 13x^2 + 4 = 0
Let's substitute y = x^2 to simplify the equation:
9y^2 - 13y + 4 = 0
Now we have a quadratic equation that we can solve using factoring, completing the square, or the quadratic formula. In this case, factoring can be a simple approach. We can find two numbers that multiply to give 36 (9 * 4) and add to give -13:
(9y - 4)(y - 1) = 0
Setting each factor equal to zero, we have:
9y - 4 = 0 or y - 1 = 0
Solving each equation for y:
9y = 4 or y = 1
Dividing both sides of the first equation by 9:
y = 4/9
Now, we substitute back for y:
x^2 = 4/9
To solve for x, we take the square root of both sides. Since we have a quadratic equation, there are usually two possible solutions:
x = ±√(4/9)
Simplifying:
x = ±(2/3)
Therefore, the solution to the first equation is x = 2/3 or x = -2/3.
To solve the second equation, x^6 - 19x^3 = 216, we notice that it can be rewritten as:
x^6 - 19x^3 - 216 = 0
We can factor the equation if we treat it as a quadratic equation with x^3 as the variable:
(x^3 - 27)(x^3 + 8) = 0
Setting each factor equal to zero, we have:
x^3 - 27 = 0 or x^3 + 8 = 0
Solving each equation for x:
x^3 = 27 or x^3 = -8
Taking the cube root of each side:
x = 3 or x = -2
Therefore, the solution to the second equation is x = 3 or x = -2.
In summary, the solutions to the given equations are:
For 9x^4 + 4 = 13x^2: x = 2/3 or x = -2/3
For x^6 - 19x^3 = 216: x = 3 or x = -2