How do I solve x^3 + x^2 -9x -9=0 for x?
I would use factoring by grouping. Factor x^2 from the first two terms and factor -9 from the last two terms. x^2 (x+1) - 9 (x+1) = 0 (x + 1) (x^2 - 9) = 0 (x + 1 ) (x + 3) (x - 3) = 0 By the zero product property, x = -1 or x = -3 or x = 3. Good luck to you !
How do I solve for x in x^3 - 3x^2 - 9x + 11?
The usual approach in a simple cubic is to try for some values. In this case the only integer values would be 1 and 11, as in factors x + 1 or x + 11. Try x + 1 or x - 1....... If x = 1, x^3 - 3x^2 - 9x + 11 = 1 - 3- 9 + 11 = 0, so x - 1 is a factor Divide the cubic by x -1 to get a quadratic and solve that to find the values you cite. Since 2x^2 + kx + 2k = 0 is a quadratic and has exactly one solution, that means it has 2 solutions that are identical. This is the case for the standard quadratic ax^2 + bx +c when b^2 - 4ac = 0 or b^2 = 4ac In the instant case,that means k^2 = 16k and k must therefore be 16 2x^2 + 16x + 32 = 0 x^2 + 8x + 16 = 0 ( a square function) (x + 4)(x + 4) = 0.......two identical solutions x = - 4
How do you solve 9^x-3^(x+1)-54=0?
First, you need to use the properties of exponents to rewrite 9^x as (3^x)^2, and 3^(x+1) as 3(3^x):9^x = (3^2)^x = 3^(2x) = (3^x)^2 because (a^m)^n = a^(mn)3^(x+1) = (3^x)(3^1) = (3^x)(3) = 3(3^x) because (a^m)(a^n) = a^(m+n)Then your equation becomes (3^x)^2 - 3(3^x) - 54 = 0 and is seem to be quadratic in 3^x, suggesting the substitution u = 3^x:u^2 - 3u - 54 = 0This equation can be solved by factoring:(u + 6)(u - 9) = 0u + 6 = 0 or u - 9 = 0u = -6 or u = 9Now back-substituting,3^x = -6 or 3^x = 9The first equation does not have a solution in the real number system, because 3^x cannot be negative, but the second one does:3^x = 93^x = 3^2So, x = 2
X^3 - 9x - 1 ;How would I solve it?
this is a depressed cubic equation as there is no x^2 term by rational root theorem we can check that this does not have a rational root there is a standard method for it http://www.sosmath.com/algebra/factor/fa... however it is too complicated to remember the steps we assume that is has gor 3 real roots and if we find later then we shall correct it now as we know cos 3t = 4 cos^3 t - 3 cos t so can we combine x^3-9x to the form some thing of this form say x = at => a^3= a^3t^3 and so we get (a is positive) f((x) = a^3t^3- 9at - 1 the ratio of coefficient of t^3 and t should be 4/3 so a^3/(9a) = 4/3 or a^2 = 12 so x = 2 sqrt(3) t so we get 24sqrt(3)t^3 - 18 sqrt(3) t - 1 = 0 so 6sqrt(3)(4t^3 - 3 t) = 1 if t = cos y then sw get 6sqrt(3) ( cos 3y) = 1 or cos 3y = 1/(6sqrt(3)) which is possible so all roots are real if cos 3y > 1 then it would not have 3 real roots and a different method would have been tried so y = cos^-1(1/(6sqrt(3)) or cos^-1(1/(6sqrt(3)) + 2pi/3 or cos^-1(1/(6sqrt(3)) + 4pi/3 so x = cos( 1/3 cos^-1(1/(6sqrt(3))) or cos (1/3 ( cos^-1(1/(6sqrt(3)) + 2pi/3)) or cos(1/3 cos^-1(1/(6sqrt(3)) + 2pi/3 ) the general method is http://mathserver.sdu.edu.cn/mathency/ma...
How to solve 3^(2x+1) - 28(3^x) + 9 = 0?
3^(2x+1)-28(3^x)+9=0 3^(2x).3 -28 (3^x)+9=0 3. (3^x)^2 -28 (3^x) + 9 = 0 Let 3^x be 'A'. Therefore, 3 A^2 - 28A + 9 = 0 ( A - 9 ) ( 3A - 1 ) = 0 A - 9 = 0 or 3A -1 = 0 A = 9 or A = 1/3 3^x = 9 or 3^x = 1/3 3^x = 3^2 or 3^x = 3^-1 x= 2 or x = -1 That's it. Hope you get it :)
How do I solve: log(base 3)X + log (base 9)X+42 = 0?
log (base 9) x = [log (base 3) x]/[log (base 3) 9] = [log (base 3) x]/2 So the equation above becomes: (3/2)[log (base 3) x] = -42 log (base 3) x = -28 x = 3^-28 = a very small number.
Lim x ---> 9 so ( x^(2) - 81 ) / ( (√x) - 3 ) solve limit?
Lim x→9 { (x² - 81) / (√x - 3) } = Lim x→9 { (x + 9)(x - 9) / (√x - 3) } = Lim x→9 { (x + 9)(√x + 3)(√x - 3) / (√x - 3) } = Lim x→9 (x + 9)(√x + 3) = (9 + 9)(3 + 3) = 18*6 = 108
How do we solve [math]x^3 - 9x^2 +54 = 0[/math] without using a calculator or guessing the first solution and then using long division?
For any cubic equation of the formf=x^3 +b .x^2 +cx+d=0Then we are looking for roots r,s,tSuch that f=(x-r)(x-s)(x-t)Expanding and equating powersWe haver+s+t =-brst=-dIn this equationx^3 -9x^2 +54=0r+s+t=9rst=-54Dot . Means multiply hereLooking at the prime factorisation of -54 =-2.3^3We might guess for a factor say t of 3Another clue is that many such problems choose the point of inflection as a root.The point of inflection ( at f''=0 is at x=-b/3 I.e. x=3) .And substituting we find that works. 27- 9.9 -54=0Sor+s= 6 ...(1)rs= -18 ...(2)(1).r -(2) givesr^2 + +rs -rs = 6r+18r^2 -6r -18 =0Using quadratic formula r,s = 3(1+-sqrt (3))The use of this technique eliminates the need for algebraic division.And the quadratic equation above is identical to that obtained by dividing f by (x-3)But you still need one root to get started and trial and error substitution is quicker than trying algebraic division.PSIf the point of inflection is a root, then it is the average of the other two rootsThus,If t is a root AND the inflection, then the other roots are att+-sqrt(t^2+d/t) or t(1+-sqrt(1+d/t^3)