Solve for x
x=1
x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333
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a+b=-7 ab=3\times 4=12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
-1,-12 -2,-6 -3,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.
-1-12=-13 -2-6=-8 -3-4=-7
Calculate the sum for each pair.
a=-4 b=-3
The solution is the pair that gives sum -7.
\left(3x^{2}-4x\right)+\left(-3x+4\right)
Rewrite 3x^{2}-7x+4 as \left(3x^{2}-4x\right)+\left(-3x+4\right).
x\left(3x-4\right)-\left(3x-4\right)
Factor out x in the first and -1 in the second group.
\left(3x-4\right)\left(x-1\right)
Factor out common term 3x-4 by using distributive property.
x=\frac{4}{3} x=1
To find equation solutions, solve 3x-4=0 and x-1=0.
3x^{2}-7x+4=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 4}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 4}}{2\times 3}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-12\times 4}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 3}
Multiply -12 times 4.
x=\frac{-\left(-7\right)±\sqrt{1}}{2\times 3}
Add 49 to -48.
x=\frac{-\left(-7\right)±1}{2\times 3}
Take the square root of 1.
x=\frac{7±1}{2\times 3}
The opposite of -7 is 7.
x=\frac{7±1}{6}
Multiply 2 times 3.
x=\frac{8}{6}
Now solve the equation x=\frac{7±1}{6} when ± is plus. Add 7 to 1.
x=\frac{4}{3}
Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.
x=\frac{6}{6}
Now solve the equation x=\frac{7±1}{6} when ± is minus. Subtract 1 from 7.
x=1
Divide 6 by 6.
x=\frac{4}{3} x=1
The equation is now solved.
3x^{2}-7x+4=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.
3x^{2}-7x+4-4=-4
Subtract 4 from both sides of the equation.
3x^{2}-7x=-4
Subtracting 4 from itself leaves 0.
\frac{3x^{2}-7x}{3}=-\frac{4}{3}
Divide both sides by 3.
x^{2}-\frac{7}{3}x=-\frac{4}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{4}{3}+\left(-\frac{7}{6}\right)^{2}
Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{4}{3}+\frac{49}{36}
Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{1}{36}
Add -\frac{4}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{7}{6}\right)^{2}=\frac{1}{36}
Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. 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{7}{6}\right)^{2}}=\sqrt{\frac{1}{36}}
Take the square root of both sides of the equation.
x-\frac{7}{6}=\frac{1}{6} x-\frac{7}{6}=-\frac{1}{6}
Simplify.
x=\frac{4}{3} x=1
Add \frac{7}{6} to both sides of the equation.
x ^ 2 -\frac{7}{3}x +\frac{4}{3} = 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 3
r + s = \frac{7}{3} rs = \frac{4}{3}
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{7}{6} - u s = \frac{7}{6} + u
Two numbers r and s sum up to \frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{3} = \frac{7}{6}. 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{7}{6} - u) (\frac{7}{6} + u) = \frac{4}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{3}
\frac{49}{36} - u^2 = \frac{4}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{4}{3}-\frac{49}{36} = -\frac{1}{36}
Simplify the expression by subtracting \frac{49}{36} on both sides
u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{6} - \frac{1}{6} = 1.000 s = \frac{7}{6} + \frac{1}{6} = 1.333
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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