Solve for x
x=-\frac{1}{2}=-0.5
x=\frac{3}{8}=0.375
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a+b=2 ab=16\left(-3\right)=-48
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
-1,48 -2,24 -3,16 -4,12 -6,8
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -48.
-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2
Calculate the sum for each pair.
a=-6 b=8
The solution is the pair that gives sum 2.
\left(16x^{2}-6x\right)+\left(8x-3\right)
Rewrite 16x^{2}+2x-3 as \left(16x^{2}-6x\right)+\left(8x-3\right).
2x\left(8x-3\right)+8x-3
Factor out 2x in 16x^{2}-6x.
\left(8x-3\right)\left(2x+1\right)
Factor out common term 8x-3 by using distributive property.
x=\frac{3}{8} x=-\frac{1}{2}
To find equation solutions, solve 8x-3=0 and 2x+1=0.
16x^{2}+2x-3=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{-2±\sqrt{2^{2}-4\times 16\left(-3\right)}}{2\times 16}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 16\left(-3\right)}}{2\times 16}
Square 2.
x=\frac{-2±\sqrt{4-64\left(-3\right)}}{2\times 16}
Multiply -4 times 16.
x=\frac{-2±\sqrt{4+192}}{2\times 16}
Multiply -64 times -3.
x=\frac{-2±\sqrt{196}}{2\times 16}
Add 4 to 192.
x=\frac{-2±14}{2\times 16}
Take the square root of 196.
x=\frac{-2±14}{32}
Multiply 2 times 16.
x=\frac{12}{32}
Now solve the equation x=\frac{-2±14}{32} when ± is plus. Add -2 to 14.
x=\frac{3}{8}
Reduce the fraction \frac{12}{32} to lowest terms by extracting and canceling out 4.
x=-\frac{16}{32}
Now solve the equation x=\frac{-2±14}{32} when ± is minus. Subtract 14 from -2.
x=-\frac{1}{2}
Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.
x=\frac{3}{8} x=-\frac{1}{2}
The equation is now solved.
16x^{2}+2x-3=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}+2x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
16x^{2}+2x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
16x^{2}+2x=3
Subtract -3 from 0.
\frac{16x^{2}+2x}{16}=\frac{3}{16}
Divide both sides by 16.
x^{2}+\frac{2}{16}x=\frac{3}{16}
Dividing by 16 undoes the multiplication by 16.
x^{2}+\frac{1}{8}x=\frac{3}{16}
Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\frac{3}{16}+\left(\frac{1}{16}\right)^{2}
Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{3}{16}+\frac{1}{256}
Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{49}{256}
Add \frac{3}{16} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{16}\right)^{2}=\frac{49}{256}
Factor x^{2}+\frac{1}{8}x+\frac{1}{256}. 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+\frac{1}{16}\right)^{2}}=\sqrt{\frac{49}{256}}
Take the square root of both sides of the equation.
x+\frac{1}{16}=\frac{7}{16} x+\frac{1}{16}=-\frac{7}{16}
Simplify.
x=\frac{3}{8} x=-\frac{1}{2}
Subtract \frac{1}{16} from both sides of the equation.
x ^ 2 +\frac{1}{8}x -\frac{3}{16} = 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 = -\frac{1}{8} rs = -\frac{3}{16}
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 = -\frac{1}{16} - u s = -\frac{1}{16} + u
Two numbers r and s sum up to -\frac{1}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{8} = -\frac{1}{16}. 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>
(-\frac{1}{16} - u) (-\frac{1}{16} + u) = -\frac{3}{16}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{16}
\frac{1}{256} - u^2 = -\frac{3}{16}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{16}-\frac{1}{256} = -\frac{49}{256}
Simplify the expression by subtracting \frac{1}{256} on both sides
u^2 = \frac{49}{256} u = \pm\sqrt{\frac{49}{256}} = \pm \frac{7}{16}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{16} - \frac{7}{16} = -0.500 s = -\frac{1}{16} + \frac{7}{16} = 0.375
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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