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How many real solutions are there to the equation 4x2 plus x 1 equals 0?
If you mean: 4x2 + x - 1 = 0 then: 4x2 + x = 1 x2 + x/4 = 1/4 x2 + x/4 + 1/64 = 17/64 (x + 1/2)2 = 17/64 x + 1/2 = ± √17 / 8 x = (-4 ± √17) / 8 Giving us two real solutions. If on the other hand you mean: 4x2 + x + 1 = 0 Then: 4x2 + x = -1 x2 + x/4 = -1/4 x2 + x/4 + 1/64 = 15/64 (x + 1/2)2 = 15/64 x + 1/2 = ± √15 / 8 x = (-4 ± √15) / 8 Again, giving us two real solutions.
What are the x-solutions for the equation 10x2-64 equals 36 plus 6x2?
If: 10x2-64 = 36+6x2 Then: 4x2-100 = 0 And: (2x-10)(2x+10) = 0 So: x = 5 or x = -5
What are the x solutions for 10x2-64 equals 36x plus 6x2?
10x2 - 64 = 36x + 6x2Subtract 6x2 from both sides:4x2 - 64 = 36xSubtract 36x from both sides:4x2 - 36x - 64 = 0Divide everything by 4:x2 - 9x - 16 = 0Use the quadratic formula:x = (9±√(92-4(-16)))/(2)x = (9-√(145))/2 or x = (9+√(145))/2
How do you solve the quadratic equation -4y2 32y-64?
Do you mean -4y2+32y-64 = 0 otherwise it's not an equation because there's no equal sign If so then by using the quadratic equation formula the values of y both equal 4
How many solutions 4x2 plus 8x plus 3 equals 0?
Start with the equation 4x2 + 8x + 3 = 0. Invoke the quadratic equation: x = [-b ± √(b2 - 4ac)]/2a, where a = 4, b = 8, and c = 3. Plug in the values for a, b, and c and solve: x = [-8 ± √(82 - 4*4*3)]/2*4, x = [-8 ± √(64 - 48)]/8, x = -1 ± √(16)/8, x = -1 ± 1/2, x = {-1/2, -3/2}.
