Skip to main content
Factor
Tick mark Image
Evaluate
Tick mark Image
Graph

Similar Problems from Web Search

Share

4x^{2}-10x+3=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 4\times 3}}{2\times 4}
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(-10\right)±\sqrt{100-4\times 4\times 3}}{2\times 4}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-16\times 3}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-10\right)±\sqrt{100-48}}{2\times 4}
Multiply -16 times 3.
x=\frac{-\left(-10\right)±\sqrt{52}}{2\times 4}
Add 100 to -48.
x=\frac{-\left(-10\right)±2\sqrt{13}}{2\times 4}
Take the square root of 52.
x=\frac{10±2\sqrt{13}}{2\times 4}
The opposite of -10 is 10.
x=\frac{10±2\sqrt{13}}{8}
Multiply 2 times 4.
x=\frac{2\sqrt{13}+10}{8}
Now solve the equation x=\frac{10±2\sqrt{13}}{8} when ± is plus. Add 10 to 2\sqrt{13}.
x=\frac{\sqrt{13}+5}{4}
Divide 10+2\sqrt{13} by 8.
x=\frac{10-2\sqrt{13}}{8}
Now solve the equation x=\frac{10±2\sqrt{13}}{8} when ± is minus. Subtract 2\sqrt{13} from 10.
x=\frac{5-\sqrt{13}}{4}
Divide 10-2\sqrt{13} by 8.
4x^{2}-10x+3=4\left(x-\frac{\sqrt{13}+5}{4}\right)\left(x-\frac{5-\sqrt{13}}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{13}}{4} for x_{1} and \frac{5-\sqrt{13}}{4} for x_{2}.
x ^ 2 -\frac{5}{2}x +\frac{3}{4} = 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 4
r + s = \frac{5}{2} rs = \frac{3}{4}
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{5}{4} - u s = \frac{5}{4} + u
Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{4}. 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{5}{4} - u) (\frac{5}{4} + u) = \frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}
\frac{25}{16} - u^2 = \frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{4}-\frac{25}{16} = -\frac{13}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = \frac{13}{16} u = \pm\sqrt{\frac{13}{16}} = \pm \frac{\sqrt{13}}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{4} - \frac{\sqrt{13}}{4} = 0.349 s = \frac{5}{4} + \frac{\sqrt{13}}{4} = 2.151
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.