what is the discriminant of the equation x^ 2 + 4x - n^2= 0 ? And how many solutions does it have?
478133
Discriminat for ax^2 + bx + c is b^2 - 4ac.\n\nThus, in our case,\n\nDiscriminat = 4^2 - 4*(1)(-n^2)\n            = 16 + 4n^2\n\nNow, since n^2 is nonnegative for any real n,\nThus, 16 + 4n^2 > 0 for all real values of n.\n\nHence, Discriminant > 0.\n\nThis means our quadratic equation above has two didtinct roots. or you can say two distinct solutions.\n\nRemark.\nFor katana's solution below, the error is that the discrminant is NOT defined to be the square root of \nb^2 -4ac but rather its just b^2-4ac \n\nRegards
227387
D = 36 - (4)*(-2)*(-3)= 36- 24 = 12\nTwo real solutions.
757145
Lets see what values we can have for f(x). Suppose f(x) = c has some solution for x.\nThen (x^2 - 1) = c(x+2), so x^2 - cx - (2c+1) = 0. For that to have a solution, we need the discriminant, (b^2-4ac, though note thats a different c!) to be at least 0. So we need c^2+4(2c+1) >= 0.\nSo c^2 + 8c + 1 >= 0.\nNow, we can write that as (c+4)^2 - 15 >= 0, so (c+4)^2 >= 15.\nThus c+4 > sqrt(15) or c+4 < -sqrt(15).\nSo c > sqrt(15)-4 or c <-sqrt(15)-4.\n\nSo f(x) > sqrt(15)-4 or f(x) < -sqrt(15) - 4.\n\n(Note you could have done the last step solving the 'c' inequality with the quadratic formula; they're both the same, really.)\n\nNote that it is a much better idea to leave the inequality until as late as possible. Note that in the above idea of gp4rts, although it may be made to work, you *can't* do the following:\n(x^2-1)/(x+2) >= a\nTherefore\n(x^2-1) >= a(x+2).\nIf you multiply by a negative, you have to switch the inequality. Thus you would have to divide everything into two cases - if x+2>0, or x+2<0, and it gets very messy.
66565
You cannot "solve" this because it is not an equation.  An equation has two parts that are joined by the symbol =.  However, in early pre-Algebra classes, they might ask you to simplify this, meaning, to write this expression so that the "x" only appears once with its coefficient and the rest of the numerical values are also combined.  For this, you add all of the coefficients of x (3+1 - invisible coefficient before the second appearance of x) to find that you have 4x.  Next, you add the -8 and 4, which you can then express as either adding negative four or as subtracting four (if you need help here, consider the number line, move eight spaces to the left of zero, then move four spaces to the right).  Thus, you can re-write the expression as 4x-4.  Look in your textbook before the problems for a better example of what exactly they are expecting you to do in that section.
402617
No solutions.\nFrom the first equation you have\ny=x^2+2x+9/2\n\nUsing the second:\nx^2+2x+3=x^2+2x+9/2\n3=9/2 , so there is no solution
970180
Use the slide method!\n2x^2 + 5x + 12\nMultiply the coefficient of x^2 with the the last term (the constant C), so it becomes:\nx^2 + 5x + 24\n(x + ) (x + )\nSince they're both positive, there are no factors of 24 you can plug into that would add up to 5.  Thus, you can't factor it.\n\n-4x^2 -9x - 10\nAgain, multiply the coefficient of x^2 with the constant.\nx^2 - 9x + 40 (it becomes positive because -10 * -4 = -40)\nThe plus sign after the 9x tells you that the signs are both going to be the same, and the subtraction sign after the x^2 tells you that they're both going to be subtraction signs.\n(x- )(x- )\nThen, think of what factors of 40 add up to 9.  None, so this can't be factored either.\n\n4a^2 - a - 5\nOnce again, multiply the coefficient of a^2 with the constant.\na^2 - a - 20\nYou can see from the - sign after the a that the signs will be different.\n(a+ ) (a- )\nThen think of factors of 20 that will equal -1.  \n(a+4) (a-5)\nFor it  to be -1, the bigger number has to be negative, so it will be a-5.\nBut then we have the problem of the -4 (coefficient of a^2) we got rid of earlier.  So, we have to divide the numbers by -4:\n(a + (4/-4))(a - (5/-4))\n(a + (4/4)) = (a - 1)\n(a - 5/-44)) doesn't divide cleanly, so the denominator goes in front of a. Thus:\n(a - 5/-4)) = (4a + 5)\n\nYes, no solution is definitely possible.  And you can use the discriminant to check:\nb^2 - 4 * a * c\na is the coefficient of x^2, b is the coefficient of x, and c is the constant.  If you plug those into the equation and you come up with a negative number, then there is no solution.\n\nYou should be able to do the last one.\nBasic rules of the slide method:\n-Multiply the coefficient of x^2 with the constant\n-Factor\n-Divide constants by coefficient: if it doesn't divide cleanly, simplify if possible and put whatever is left of denominator in front of the x.  For example:\n(x + 4/6) = (3x + 2) (because you can simplify both 4 and 6 by dividing by 2)
516234
The discriminant = b^2-4ac = 1 - 4.2.(-4) = 33 > 0 \n⇒ We can get two unequal, real but irrational roots.\n\nThis is so, because, by actually calculating the roots of the equation applying formula. we find that:\nx= [1±√33]/4
641623
(-6)^2 - 4(9)(1) = 0\n\nThe discriminant is b^2 - 4ac, where a, b, and c are such that\n\nax^2 + bx + c = 0\n\nFor your formula, 9x^2 - 6x + 1 = 0, so a = 9, b = -6, and c = 1\n\nSince the discriminant is zero, there is only one real root to this equation.  If the discriminant is positive, then the quadratic has two real roots.  If the discriminant is negativ, then there are no real roots (both roots are complex).  This only has one root because\n\n9x^2-6x+1 = (3x - 1)^2 = 0.\n\nSo 3x-1 = 0.\nSo 3x = 1.\nSo x = 1/3.
992331
The trick here is to note that it looks like a quadratic formula.  Let's do a little manuevering to see this:\n\nx^4 = (x^2)^2\n\nwhich means that we can re-write the problem as (x^2)^2 + x^2 + 1.  Now, substitute another letter, say t for x^2.  We get:\n\nt^2 + t + 1\n\nAll I've done is replace the two x^2 values with 't'.  Now, it's just a quadratic equation of the form:\n\nax^2 + bx + c\n\nTo factor this, you'll need to find two values that multiply to the last term (1, in this case) and add to the coefficient of the second term (which is also 1, in this case).  \n\nWith a little investigation, you'll find that two integers do not exist with such properties, which for most math books, means that, though the problem has a solution, it can't be factored.\n\n***THE ANSWER STOPS HERE if you haven't learned about imaginary numbers.***\n\n*However*, in case you really want to factor this, you simply use the quadratic formula to find the values of x that solve this equation:\n\nt = (-b +/- sqrt(b^2 - 4ac))/2a\n\nSo, we get that t = (-1 + sqrt(-3))/2 and t = (-1 - sqrt(-3))/2, which are imaginary numbers.  These two solutions I will call solutionone and solutiontwo.  \n\nSo the factorization for f(t) is:\n\n(t - solutionone)(t - solutiontwo) where solutionone and solutiontwo are the values that come out of the quadratic formula:\n\nThen, you plug x^2 back in for t and get:\n\n(x^2 - solutionone)(x^2 - solutiontwo)\n\nThese need to be factored as well (ugh) because a linear factorization implies that all the factors have x instead of x^2.  \n\nThank goodness, though, that these are really easy.\n\nIt is always the case that x^2 - a can be factored like this:\n\n(x - sqrt(a))(x + sqrt(a)).\n\nAlways.\n\nSo, this means that the complete factorization for the problem is:\n\n(x - sqrt(solutionone))(x + sqrt(solutionone))(x - sqrt(solutiontwo))(x + sqrt(solutiontwo))\n\nYeow!  Good luck!
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
859279
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
695292
Put it into standard form and factor or use the quadratic formula.\n\nYou want the discriminate to be positive\nin standard form b=coeff of x: 2k-8\na = k\nc = k\n\nnow plug and chug and find values of k that make the discriminant positive\nb^2 - 4ac =
904373
Let y = x^2\n\nSo you have y^2 + 3y - 2\nYour discriminant (b² - 4ac) = 3² - 4*1*(-2) = 17\n\nSo that means you have two real roots for y^2\ny = x^2 = (-3 +/- sqrt(17))/2\ny = x^2 = 0.561552813 or -3.56155281\n\nWhen you solve for x, you have:\nx = sqrt(0.561552813) or sqrt( -3.56155281)\n\nSo you'll have two real roots:\nx = +/- 0.749368276\n\nAnd two imaginary roots:\nx = +/- 1.88720768 i
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
220199
The problem factors to\n\n(x - 2)(x^2 + 1)(x^2 + 1) or (x - 2)(x^2 + 1)^2\n\nx = 2, i, -i\n\nthe only real zero is 2.
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
662920
The discriminant, b^2 - 4ac, is part of the "quadratic formula", which solves any quadratic equation.\n\nThe reason the discriminant gives information about the nature of the solutions is that in the quadratic formula, it is square rooted.\n\nNow, if the discriminant is negative, then the solutions are non-real, since you are taking the square root of a negative number.  In other words, there are no (real) solutions.\n\nIf the discriminant is equal to zero, then the square rooted part of the equation goes away completely, leaving only one solution (or, more precisely, two equal solutions), equal to -b/2a.\n\nIf the discriminant is positive, then you are taking the square root of a positive number, yeilding two real (and distinct) solutions.  (There are two solutions because of the "plus and minus" in front of the square root sign.)\n\nIn the special case where the discriminant is positive AND a perfect square, then the two solutions will be rational, since the square root will be a whole number.  If the discriminant is positive and not a perfect square, then the two solutions will be irrational (ie. have a square root sign in them).