Solve for x
x=1
x=9
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16x^{2}-160x+144=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-160\right)±\sqrt{\left(-160\right)^{2}-4\times 16\times 144}}{2\times 16}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -160 for b, and 144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-160\right)±\sqrt{25600-4\times 16\times 144}}{2\times 16}
Square -160.
x=\frac{-\left(-160\right)±\sqrt{25600-64\times 144}}{2\times 16}
Multiply -4 times 16.
x=\frac{-\left(-160\right)±\sqrt{25600-9216}}{2\times 16}
Multiply -64 times 144.
x=\frac{-\left(-160\right)±\sqrt{16384}}{2\times 16}
Add 25600 to -9216.
x=\frac{-\left(-160\right)±128}{2\times 16}
Take the square root of 16384.
x=\frac{160±128}{2\times 16}
The opposite of -160 is 160.
x=\frac{160±128}{32}
Multiply 2 times 16.
x=\frac{288}{32}
Now solve the equation x=\frac{160±128}{32} when ± is plus. Add 160 to 128.
x=9
Divide 288 by 32.
x=\frac{32}{32}
Now solve the equation x=\frac{160±128}{32} when ± is minus. Subtract 128 from 160.
x=1
Divide 32 by 32.
x=9 x=1
The equation is now solved.
16x^{2}-160x+144=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
16x^{2}-160x+144-144=-144
Subtract 144 from both sides of the equation.
16x^{2}-160x=-144
Subtracting 144 from itself leaves 0.
\frac{16x^{2}-160x}{16}=-\frac{144}{16}
Divide both sides by 16.
x^{2}+\left(-\frac{160}{16}\right)x=-\frac{144}{16}
Dividing by 16 undoes the multiplication by 16.
x^{2}-10x=-\frac{144}{16}
Divide -160 by 16.
x^{2}-10x=-9
Divide -144 by 16.
x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-10x+25=-9+25
Square -5.
x^{2}-10x+25=16
Add -9 to 25.
\left(x-5\right)^{2}=16
Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-5\right)^{2}}=\sqrt{16}
Take the square root of both sides of the equation.
x-5=4 x-5=-4
Simplify.
x=9 x=1
Add 5 to both sides of the equation.
x ^ 2 -10x +9 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 16
r + s = 10 rs = 9
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(5 - u) (5 + u) = 9
To solve for unknown quantity u, substitute these in the product equation rs = 9
25 - u^2 = 9
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 9-25 = -16
Simplify the expression by subtracting 25 on both sides
u^2 = 16 u = \pm\sqrt{16} = \pm 4
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =5 - 4 = 1 s = 5 + 4 = 9
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}