Solve 4x^2-1=35x-70

$$4 { x }^{ 2 } -1 = 35x-70$$

Solve for x

$$x=3$$

$$x = \frac{23}{4} = 5\frac{3}{4} = 5.75$$

Steps Using Factoring By Grouping

Subtract 35x from both sides.

$$4x^{2}-1-35x=-70$$

Add 70 to both sides.

$$4x^{2}-1-35x+70=0$$

Add -1 and 70 to get 69.

$$4x^{2}+69-35x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{2}-35x+69=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+69. To find a and b, set up a system to be solved.

$$a+b=-35 ab=4\times 69=276$$

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 276.

$$-1,-276 -2,-138 -3,-92 -4,-69 -6,-46 -12,-23$$

Calculate the sum for each pair.

$$-1-276=-277 -2-138=-140 -3-92=-95 -4-69=-73 -6-46=-52 -12-23=-35$$

The solution is the pair that gives sum -35.

$$a=-23 b=-12$$

Rewrite 4x^{2}-35x+69 as \left(4x^{2}-23x\right)+\left(-12x+69\right).

$$\left(4x^{2}-23x\right)+\left(-12x+69\right)$$

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

$$x\left(4x-23\right)-3\left(4x-23\right)$$

Factor out common term 4x-23 by using distributive property.

$$\left(4x-23\right)\left(x-3\right)$$

To find equation solutions, solve 4x-23=0 and x-3=0.

$$x=\frac{23}{4} x=3$$

Steps Using the Quadratic Formula

Subtract 35x from both sides.

$$4x^{2}-1-35x=-70$$

Add 70 to both sides.

$$4x^{2}-1-35x+70=0$$

Add -1 and 70 to get 69.

$$4x^{2}+69-35x=0$$

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.

$$4x^{2}-35x+69=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -35 for b, and 69 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 4\times 69}}{2\times 4}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 4\times 69}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-35\right)±\sqrt{1225-16\times 69}}{2\times 4}$$

Multiply -16 times 69.

$$x=\frac{-\left(-35\right)±\sqrt{1225-1104}}{2\times 4}$$

Add 1225 to -1104.

$$x=\frac{-\left(-35\right)±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$x=\frac{-\left(-35\right)±11}{2\times 4}$$

The opposite of -35 is 35.

$$x=\frac{35±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{35±11}{8}$$

Now solve the equation x=\frac{35±11}{8} when ± is plus. Add 35 to 11.

$$x=\frac{46}{8}$$

Reduce the fraction \frac{46}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{23}{4}$$

Now solve the equation x=\frac{35±11}{8} when ± is minus. Subtract 11 from 35.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

The equation is now solved.

$$x=\frac{23}{4} x=3$$

Steps for Completing the Square

Subtract 35x from both sides.

$$4x^{2}-1-35x=-70$$

Add 1 to both sides.

$$4x^{2}-35x=-70+1$$

Add -70 and 1 to get -69.

$$4x^{2}-35x=-69$$

Divide both sides by 4.

$$\frac{4x^{2}-35x}{4}=-\frac{69}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{35}{4}x=-\frac{69}{4}$$

Divide -\frac{35}{4}, the coefficient of the x term, by 2 to get -\frac{35}{8}. Then add the square of -\frac{35}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{4}x+\left(-\frac{35}{8}\right)^{2}=-\frac{69}{4}+\left(-\frac{35}{8}\right)^{2}$$

Square -\frac{35}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{4}x+\frac{1225}{64}=-\frac{69}{4}+\frac{1225}{64}$$

Add -\frac{69}{4} to \frac{1225}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{35}{4}x+\frac{1225}{64}=\frac{121}{64}$$

Factor x^{2}-\frac{35}{4}x+\frac{1225}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x-\frac{35}{8}=\frac{11}{8} x-\frac{35}{8}=-\frac{11}{8}$$

Add \frac{35}{8} to both sides of the equation.

$$x=\frac{23}{4} x=3$$