Solve 52t^2+9t-1=

$$52 t ^ { 2 } + 9 t - 1 =$$

Factor

$$\left(13t-1\right)\left(4t+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 52t^{2}+at+bt-1. To find a and b, set up a system to be solved.

$$a+b=9 ab=52\left(-1\right)=-52$$

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 -52.

$$-1,52 -2,26 -4,13$$

Calculate the sum for each pair.

$$-1+52=51 -2+26=24 -4+13=9$$

The solution is the pair that gives sum 9.

$$a=-4 b=13$$

Rewrite 52t^{2}+9t-1 as \left(52t^{2}-4t\right)+\left(13t-1\right).

$$\left(52t^{2}-4t\right)+\left(13t-1\right)$$

Factor out 4t in 52t^{2}-4t.

$$4t\left(13t-1\right)+13t-1$$

Factor out common term 13t-1 by using distributive property.

$$\left(13t-1\right)\left(4t+1\right)$$

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.

$$52t^{2}+9t-1=0$$

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.

$$t=\frac{-9±\sqrt{9^{2}-4\times 52\left(-1\right)}}{2\times 52}$$

Square 9.

$$t=\frac{-9±\sqrt{81-4\times 52\left(-1\right)}}{2\times 52}$$

Multiply -4 times 52.

$$t=\frac{-9±\sqrt{81-208\left(-1\right)}}{2\times 52}$$

Multiply -208 times -1.

$$t=\frac{-9±\sqrt{81+208}}{2\times 52}$$

Add 81 to 208.

$$t=\frac{-9±\sqrt{289}}{2\times 52}$$

Take the square root of 289.

$$t=\frac{-9±17}{2\times 52}$$

Multiply 2 times 52.

$$t=\frac{-9±17}{104}$$

Now solve the equation t=\frac{-9±17}{104} when ± is plus. Add -9 to 17.

$$t=\frac{8}{104}$$

Reduce the fraction \frac{8}{104} to lowest terms by extracting and canceling out 8.

$$t=\frac{1}{13}$$

Now solve the equation t=\frac{-9±17}{104} when ± is minus. Subtract 17 from -9.

$$t=-\frac{26}{104}$$

Reduce the fraction \frac{-26}{104} to lowest terms by extracting and canceling out 26.

$$t=-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{13} for x_{1} and -\frac{1}{4} for x_{2}.

$$52t^{2}+9t-1=52\left(t-\frac{1}{13}\right)\left(t-\left(-\frac{1}{4}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$52t^{2}+9t-1=52\left(t-\frac{1}{13}\right)\left(t+\frac{1}{4}\right)$$

Subtract \frac{1}{13} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$52t^{2}+9t-1=52\times \frac{13t-1}{13}\left(t+\frac{1}{4}\right)$$

Add \frac{1}{4} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$52t^{2}+9t-1=52\times \frac{13t-1}{13}\times \frac{4t+1}{4}$$

Multiply \frac{13t-1}{13} times \frac{4t+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$52t^{2}+9t-1=52\times \frac{\left(13t-1\right)\left(4t+1\right)}{13\times 4}$$

Multiply 13 times 4.

$$52t^{2}+9t-1=52\times \frac{\left(13t-1\right)\left(4t+1\right)}{52}$$

Cancel out 52, the greatest common factor in 52 and 52.

$$52t^{2}+9t-1=\left(13t-1\right)\left(4t+1\right)$$

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 52

$$x ^ 2 +\frac{9}{52}x -\frac{1}{52} = 0$$

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

$$r + s = -\frac{9}{52} rs = -\frac{1}{52}$$

Two numbers r and s sum up to -\frac{9}{52} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{52} = -\frac{9}{104}. 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>

$$r = -\frac{9}{104} - u s = -\frac{9}{104} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{52}

$$(-\frac{9}{104} - u) (-\frac{9}{104} + u) = -\frac{1}{52}$$

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

$$\frac{81}{10816} - u^2 = -\frac{1}{52}$$

Simplify the expression by subtracting \frac{81}{10816} on both sides

$$-u^2 = -\frac{1}{52}-\frac{81}{10816} = -\frac{289}{10816}$$

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

$$u^2 = \frac{289}{10816} u = \pm\sqrt{\frac{289}{10816}} = \pm \frac{17}{104}$$

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

$$r =-\frac{9}{104} - \frac{17}{104} = -0.250 s = -\frac{9}{104} + \frac{17}{104} = 0.077$$

Evaluate

$$\left(13t-1\right)\left(4t+1\right)$$