Solve 49x^2-77x+30

$$49 x ^ { 2 } - 77 x + 30$$

Factor

$$\left(7x-6\right)\left(7x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 49x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-77 ab=49\times 30=1470$$

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 1470.

$$-1,-1470 -2,-735 -3,-490 -5,-294 -6,-245 -7,-210 -10,-147 -14,-105 -15,-98 -21,-70 -30,-49 -35,-42$$

Calculate the sum for each pair.

$$-1-1470=-1471 -2-735=-737 -3-490=-493 -5-294=-299 -6-245=-251 -7-210=-217 -10-147=-157 -14-105=-119 -15-98=-113 -21-70=-91 -30-49=-79 -35-42=-77$$

The solution is the pair that gives sum -77.

$$a=-42 b=-35$$

Rewrite 49x^{2}-77x+30 as \left(49x^{2}-42x\right)+\left(-35x+30\right).

$$\left(49x^{2}-42x\right)+\left(-35x+30\right)$$

Factor out 7x in the first and -5 in the second group.

$$7x\left(7x-6\right)-5\left(7x-6\right)$$

Factor out common term 7x-6 by using distributive property.

$$\left(7x-6\right)\left(7x-5\right)$$

Steps Using the Quadratic Formula

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.

$$49x^{2}-77x+30=0$$

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.

$$x=\frac{-\left(-77\right)±\sqrt{\left(-77\right)^{2}-4\times 49\times 30}}{2\times 49}$$

Square -77.

$$x=\frac{-\left(-77\right)±\sqrt{5929-4\times 49\times 30}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-77\right)±\sqrt{5929-196\times 30}}{2\times 49}$$

Multiply -196 times 30.

$$x=\frac{-\left(-77\right)±\sqrt{5929-5880}}{2\times 49}$$

Add 5929 to -5880.

$$x=\frac{-\left(-77\right)±\sqrt{49}}{2\times 49}$$

Take the square root of 49.

$$x=\frac{-\left(-77\right)±7}{2\times 49}$$

The opposite of -77 is 77.

$$x=\frac{77±7}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{77±7}{98}$$

Now solve the equation x=\frac{77±7}{98} when ± is plus. Add 77 to 7.

$$x=\frac{84}{98}$$

Reduce the fraction \frac{84}{98} to lowest terms by extracting and canceling out 14.

$$x=\frac{6}{7}$$

Now solve the equation x=\frac{77±7}{98} when ± is minus. Subtract 7 from 77.

$$x=\frac{70}{98}$$

Reduce the fraction \frac{70}{98} to lowest terms by extracting and canceling out 14.

$$x=\frac{5}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{7} for x_{1} and \frac{5}{7} for x_{2}.

$$49x^{2}-77x+30=49\left(x-\frac{6}{7}\right)\left(x-\frac{5}{7}\right)$$

Subtract \frac{6}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$49x^{2}-77x+30=49\times \frac{7x-6}{7}\left(x-\frac{5}{7}\right)$$

Subtract \frac{5}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$49x^{2}-77x+30=49\times \frac{7x-6}{7}\times \frac{7x-5}{7}$$

Multiply \frac{7x-6}{7} times \frac{7x-5}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$49x^{2}-77x+30=49\times \frac{\left(7x-6\right)\left(7x-5\right)}{7\times 7}$$

Multiply 7 times 7.

$$49x^{2}-77x+30=49\times \frac{\left(7x-6\right)\left(7x-5\right)}{49}$$

Cancel out 49, the greatest common factor in 49 and 49.

$$49x^{2}-77x+30=\left(7x-6\right)\left(7x-5\right)$$

Steps Using Direct Factoring Method

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 49

$$x ^ 2 -\frac{11}{7}x +\frac{30}{49} = 0$$

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

$$r + s = \frac{11}{7} rs = \frac{30}{49}$$

Two numbers r and s sum up to \frac{11}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{7} = \frac{11}{14}. 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>

$$r = \frac{11}{14} - u s = \frac{11}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{30}{49}

$$(\frac{11}{14} - u) (\frac{11}{14} + u) = \frac{30}{49}$$

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

$$\frac{121}{196} - u^2 = \frac{30}{49}$$

Simplify the expression by subtracting \frac{121}{196} on both sides

$$-u^2 = \frac{30}{49}-\frac{121}{196} = -\frac{1}{196}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1}{196} u = \pm\sqrt{\frac{1}{196}} = \pm \frac{1}{14}$$

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

$$r =\frac{11}{14} - \frac{1}{14} = 0.714 s = \frac{11}{14} + \frac{1}{14} = 0.857$$

Evaluate

$$\left(7x-6\right)\left(7x-5\right)$$