Solve 3x^2-x-70=0 | Microsoft Math Solver (2024)

a+b=-1 ab=3\left(-70\right)=-210

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-70. To find a and b, set up a system to be solved.

1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -210.

1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1

Calculate the sum for each pair.

a=-15 b=14

The solution is the pair that gives sum -1.

\left(3x^{2}-15x\right)+\left(14x-70\right)

Rewrite 3x^{2}-x-70 as \left(3x^{2}-15x\right)+\left(14x-70\right).

3x\left(x-5\right)+14\left(x-5\right)

Factor out 3x in the first and 14 in the second group.

\left(x-5\right)\left(3x+14\right)

Factor out common term x-5 by using distributive property.

x=5 x=-\frac{14}{3}

To find equation solutions, solve x-5=0 and 3x+14=0.

3x^{2}-x-70=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(-1\right)±\sqrt{1-4\times 3\left(-70\right)}}{2\times 3}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-1\right)±\sqrt{1-12\left(-70\right)}}{2\times 3}

Multiply -4 times 3.

x=\frac{-\left(-1\right)±\sqrt{1+840}}{2\times 3}

Multiply -12 times -70.

x=\frac{-\left(-1\right)±\sqrt{841}}{2\times 3}

Add 1 to 840.

x=\frac{-\left(-1\right)±29}{2\times 3}

Take the square root of 841.

x=\frac{1±29}{2\times 3}

The opposite of -1 is 1.

x=\frac{1±29}{6}

Multiply 2 times 3.

x=\frac{30}{6}

Now solve the equation x=\frac{1±29}{6} when ± is plus. Add 1 to 29.

x=5

Divide 30 by 6.

x=-\frac{28}{6}

Now solve the equation x=\frac{1±29}{6} when ± is minus. Subtract 29 from 1.

x=-\frac{14}{3}

Reduce the fraction \frac{-28}{6} to lowest terms by extracting and canceling out 2.

x=5 x=-\frac{14}{3}

The equation is now solved.

3x^{2}-x-70=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}-x-70-\left(-70\right)=-\left(-70\right)

Add 70 to both sides of the equation.

3x^{2}-x=-\left(-70\right)

Subtracting -70 from itself leaves 0.

3x^{2}-x=70

Subtract -70 from 0.

\frac{3x^{2}-x}{3}=\frac{70}{3}

Divide both sides by 3.

x^{2}-\frac{1}{3}x=\frac{70}{3}

Dividing by 3 undoes the multiplication by 3.

x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{70}{3}+\left(-\frac{1}{6}\right)^{2}

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{70}{3}+\frac{1}{36}

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{841}{36}

Add \frac{70}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

\left(x-\frac{1}{6}\right)^{2}=\frac{841}{36}

Factor x^{2}-\frac{1}{3}x+\frac{1}{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{1}{6}\right)^{2}}=\sqrt{\frac{841}{36}}

Take the square root of both sides of the equation.

x-\frac{1}{6}=\frac{29}{6} x-\frac{1}{6}=-\frac{29}{6}

Simplify.

x=5 x=-\frac{14}{3}

Add \frac{1}{6} to both sides of the equation.

x ^ 2 -\frac{1}{3}x -\frac{70}{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{1}{3} rs = -\frac{70}{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{1}{6} - u s = \frac{1}{6} + u

Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{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{1}{6} - u) (\frac{1}{6} + u) = -\frac{70}{3}

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{70}{3}

\frac{1}{36} - u^2 = -\frac{70}{3}

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -\frac{70}{3}-\frac{1}{36} = -\frac{841}{36}

Simplify the expression by subtracting \frac{1}{36} on both sides

u^2 = \frac{841}{36} u = \pm\sqrt{\frac{841}{36}} = \pm \frac{29}{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{1}{6} - \frac{29}{6} = -4.667 s = \frac{1}{6} + \frac{29}{6} = 5

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.