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What are the x solutions for the equation 10x2 64 equals 36 plus 6x2?

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2012-05-04 20:12:02
2012-05-04 20:12:02

If the 64 has a minus sign, then the solution is x = +/-5

Simplifying gives 4x squared = 100 ie x squared = 25...

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10x2 - 64 = 36 + 6x2; whence, 4x2 - 100 = 0, x2 = 25, and x = ±5.


If: 10x2-64 = 36+6x2 Then: 4x2-100 = 0 And: (2x-10)(2x+10) = 0 So: x = 5 or x = -5


That doesn't factor neatly. Applying the quadratic equation, we find two real solutions: (31 plus or minus the square root of 401) divided by 20x = 2.5512492197250394x = 0.5487507802749606




There are no real solutions because the discriminant of the quadratic equation is less than zero.


That equation has no solutions.There's no number that can go in place of 'x' that canmake the equation a true statement.


It is a quadratic equation and its solutions can be found by using the quadratic equation formula.


zero solutions. If you plot these two lines, you will see that they are parallel and do not intersect.


No. It's a quadratic equation, and it has two solutions.


It is a quadratic equation that has 2 solutions


10x2 - 64 = 36 + 6x2 Combine like terms: 4x2 = 100 Divide both sides by 4: x2 = 25 Take square roots: x = -5 or +5


It has no solutions because the discriminant of the quadratic equation is less than zero.



It will have two solutions because its a quadratic equation: x = -8.472135955 or x = 0.472135955


x2+7x+3 = 0 Using the quadratic equation formula the solutions are:- x = -6.541381265 or x = -0.4586187349



It can't be factored because its discriminant is less than zero.



The equation soild plus gas equals a liquid


If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7


Since the highest power of 'x' in the equation is '1', there's one solution.


10x2 - 20xz + 15xy = 5x (2x - 4z + 3y)


Strictly speaking the above equation is a tautological equation or an IDENTITY. An identity is true for all values of any variables that appear in it. Thus, the above "equation" is true for all value of x. - that is, it has infinitely many solutions.


There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis



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