Is it correct that an equation of degree 2 is called quaderatic equation?
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
293371
The equation has 0 real solutions, but 2 complex solutions.\n\nThere is a very fundamental theorem of algebra which states that any polynomial equation of degree n has n roots. In your case, the degree is 2, so the equation has to have 2 roots.\n\nHowever, roots can be either real, complex, or repeated roots. For example, x^2=0 appears to only have one solutions (x=0) but x=0 is a repeated root, occurring twice. For polynomials of degree 2, you can find out about the solutions by using the discriminant. An equation of the form ax^2+bx+c=0 has discriminant b^2-4ac. The discriminant can be one of three things:\n\n>0 This means that both roots are REAL roots.\n=0 This means that there is ONE REAL root, and it is a repeated root.\n<0 This means that both roots are COMPLEX roots and there is no real solution.\n\nThe methods for general degree n polynomials are much more difficult, if not impossible, to classify.
1157534
Tell your teacher this is a poorly worded question. ALL quadratic functions have two solutions. You could have two real solutions, one real solution that is repeated once, or two imaginary (complex) solutions.
879052
x^2 - 8x + 16\n\n\nthe person above me said it is -x^2 but that is incorrect because -x * -x = x^2
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
703053
so the Q is basically asking for two binomials which, if multiplied will end up as a polinomial...\nAs a matter of fact any two binomials can form a polynomial.\n(2X+3) (x-7)...this is an e.g.
745544
the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366
166521
Yes it is; and let us see how:\n\n(a+b)square = (a+b) x (a+b)\n= a x a + a x b + b x a + b x b\n= a x a + a x b + a x b + b x b\n= a square + 2ab + b square\n\nsimple, isnt it?
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
840916
i know how it's done. in the theorm, every number is one higher than stated (ex. 1=2, 2=3, 3=4, etc) since it's 2+2, it's actually 3+3. and if 3+3 =6, and 6=7, then the resulting equation is 2+2=7
115194
3 + 2sqrt(2) = (1 + sqrt(2))^2 so sqrt(3+2sqrt(2))=1+sqrt(2).\n\nBugzpodder, (x^2-3)^2 - 8 = x^4 - 6 x^2 + 1 is reducible.  Namely, it is x^4 - 2 x^2 + 1 - 4 x^2 = (x^2-1)^2 - (2x)^2\n   = (x^2 - 2x - 1)(x^2 + 2x - 1)
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
741426
If z=a+ib, then we have:\n2a=a²+b²\na²-2a+b²=0\nThis is a quadratic equation in a, with solutions:\n\na=(2±√(4-4b²)/2\na=1±√(1-b²)\n\nNote that since both a and b are real numbers, this implies that 0≤a≤2 and -1≤b≤1.
739088
Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.
1057648
You always can bring a quadratic equation\nax2 + bx + c = 0  in the form x2 + 2px + q = 0\nby dividing with a and p = b/(2a) while q = c/a\n\nx2 + 2px + q = 0\nx2 + 2px + p2 + q = p2\n(x + p)^2 = p2 - q\nx + p = +V(p2 - q) or x + p =  -V(p2 - q) \nx  = -p +V(p2 - q) or x = -p  - V(p2 - q)\n\nWhen you replace p and q you find the famous abc formula.\nSatisfied?
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!