If 4x^2+kx=-9, find the value of k, k>0 such that the equation has two equal roots.?
818628
4x^2 + kx + 9 = 0.\nTwo methods: a) there are two equal roots if the discriminant, which is b^2 - 4ac, equals 0.\nSo k^2 - 4*4*9 = 0, so k = 12. (since k>0)\nb) there are two equal roots if the quadratic is actually a perfect square. You need a 4 = 2^2 on the front, and a 3 = 3^2 on the end, so it must equal (2x+3)^2 (since the middle bit must be > 0).That is 4x^2 + 12x + 9, so k = 12.
784622
16x^2 - 4x - 2\n= (4x)^2 -(4x)- 2\n= {(4x)^2 - 1} - (4x+1)\n={(4x+1)(4x-1)} - (4x+1)\n=(4x+1)(4x-2)\n= 2*(4x+1)(2x-1)
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
530142
use the quadratic equation:\nx=-b +/- the square root of b^2 - 4ac all over 2a\na=1,b=7,c=10 so plug it in\nx=-7 +/- the square root of 49-40 all over 2\n-7 +3 = -4 divided by 2 x=-2\n-7-3=-10 divided by 2 = -5\nso x = -2 or x=-5
1021313
Replace x with 6\n6^2 - 2(6) + k = 0\n36 - 12 + k = 0\n24 + k = 0\nk = -24\nThe other root is -24/6 = -4
173634
(4x^2 + x + 3 ) + ( 5x^2 + 9x - 2 )\n= 4x^2 + x + 3 + 5x^2 + 9x - 2 \n= (4x^2 + 5x^2) + (x + 9x) + (3 - 2)\n= 9x^2 + 10x + 1\n= 9x^2 + 9x + x + 1\n= (9x^2 + 9x) + (x + 1)\n= 9x(x + 1) + 1(x + 1)\n= (9x+1)(x+1)\nx = -1/9 or -1\n\nEDIT: Since this is an expression, the answer is  (9x+1)(x+1). If this was an equation, the answer would be \nx = -1/9 or -1.
736851
I got a different answer than you.  Let's do this together:\n\nRecall that for an equation ax² + bx + c = 0, that the quadratic formula is:\n\nx = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\nSo we want to find the zeros of x² + 10x + 23\n\nSo, x = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\n= [ -10 ± √ (10² - 4*1*23) ] ÷ 2*1\n\n= [ -10 ± √ (100 - 92) ] ÷ 2\n\n= [ -10 ± √8 ] ÷ 2\n\n= [ -10 ± 2√2 ] ÷ 2\n\n= -5 ±√2\n\nHope that helped.\n\n~ ♥ ~
931431
Let the roots be a and b\na+b=4\nab=3\n(a-b)^2=(a+b)^2-4ab=16-12=4\na-b=2\n\na=3, b=1 or a=1, b=3\n\nthe equation has roots 3 &1\nSo the equation is: (x-3)(x-1)=0\n\ni.e. x^2-4x+3=0
672081
Answer: x = 1 or 4\n\nEXPLANATION:\nx^3 - 6x^2 + 9x - 4 = 0\nThis can be written as:\nx^3 - x^2 - 5x^2 + 5x + 4x - 4 = 0\nor x^2(x-1) - 5x(x-1) + 4(x-1) = 0\nor (x-1)(x^2 - 5x + 4) = 0\nor (x-1)(x^2 - 4x - x + 4) = 0\nor (x-1){x(x-4) - 1(x - 4)} = 0\nor {(x-1)^2}(x-4) = 0\nTherefore either x = 1 or x = 4
285238
you can take 9v^2 out and then it becomes a simple quadratic equation\ni can give u d ans \nit will be\n\n9v^2(v+2)(v+1)\ni hope u understood
369100
a(x^2)+2*(ax)+a=0\na(x^2+2x+1)\nLet a=0\nx^2+2x+1=0\n(x+1)^2=0\nsquare root of (x+1)^2=x+1 & square root of 0=0 \nSo,x+1=0\nx=0-1\nx= -1
800533
Well dear Kristi \n\nthe answer is \n( 9r - 2s )^2 \n{ as you know ( a - b )^2 = a^2 -2ab+ b^2 \n81r^2 - 36rs + 4s^2\na = 9r , b = 2s , so \na^2 = (9r)^2 = 81r^2 \nb^2 = (2s)^2 = 4s^2 \n-2ab = (-2)(9r)(2s) = -36rs  } \n\nGood Luck Darling
245342
A 4th degree equation has 4 roots^_^\nx^4 = 9\nx^2 = +3 or x^2 = -3\nx = +sqrt3 or x = +sqrt(-3)\nx = -sqrt3 or x = -sqrt(-3)\n\nTherefore:\nx = +sqrt3\nx = -sqrt3\nx = +i*sqrt3\nx = -i*sqrt3\n\n^_^
834770
3x^2-x-4=0\n=> 3x^2-4x+3x-4=0\n=> x(3x-4)+1(3x-4)=0\n=>(3x-4)(x+1)=0\nTherefore,either 3x-4=0 or x+1=0\nIf 3x-4=0,then x=4/3,and \nif x+1=0,then x=-1\nTherefore, x=4/3 or -1 ans
65662
the above answer is wrong.\n\n( x^2 + 7x - 4 ) + ( x^2 + 9 ) = x^2 + x^2 + 7x -4 +9\n= 2x^2 + 7x + 5
771896
expand x^2-6x+9 = 5\nor x^2-6x+4 =0
891094
http://mathforum.org/library/drmath/view/60548.html\n\nyour equation:\ny+9=4x-8\n\nbring x to the left side, 9 to the right side\n-4x + y = -8 - 9\n\nsimplify\n-4x + y = -17\n\nthere are two negatives; multiply through the whole thing with -1\n4x - y = 17\n\nFinal answer.