Solve 49x^2+9=42x

$$49 x ^ { 2 } + 9 = 42 x$$

Solve for x

$$x=\frac{3}{7}\approx 0.428571429$$

Steps Using Factoring By Grouping

Subtract 42x from both sides.

$$49x^{2}+9-42x=0$$

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

$$49x^{2}-42x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 49x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-42 ab=49\times 9=441$$

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 441.

$$-1,-441 -3,-147 -7,-63 -9,-49 -21,-21$$

Calculate the sum for each pair.

$$-1-441=-442 -3-147=-150 -7-63=-70 -9-49=-58 -21-21=-42$$

The solution is the pair that gives sum -42.

$$a=-21 b=-21$$

Rewrite 49x^{2}-42x+9 as \left(49x^{2}-21x\right)+\left(-21x+9\right).

$$\left(49x^{2}-21x\right)+\left(-21x+9\right)$$

Factor out 7x in the first and -3 in the second group.

$$7x\left(7x-3\right)-3\left(7x-3\right)$$

Factor out common term 7x-3 by using distributive property.

$$\left(7x-3\right)\left(7x-3\right)$$

Rewrite as a binomial square.

$$\left(7x-3\right)^{2}$$

To find equation solution, solve 7x-3=0.

$$x=\frac{3}{7}$$

Steps Using the Quadratic Formula

Subtract 42x from both sides.

$$49x^{2}+9-42x=0$$

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.

$$49x^{2}-42x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -42 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 49\times 9}}{2\times 49}$$

Square -42.

$$x=\frac{-\left(-42\right)±\sqrt{1764-4\times 49\times 9}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-42\right)±\sqrt{1764-196\times 9}}{2\times 49}$$

Multiply -196 times 9.

$$x=\frac{-\left(-42\right)±\sqrt{1764-1764}}{2\times 49}$$

Add 1764 to -1764.

$$x=\frac{-\left(-42\right)±\sqrt{0}}{2\times 49}$$

Take the square root of 0.

$$x=-\frac{-42}{2\times 49}$$

The opposite of -42 is 42.

$$x=\frac{42}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{42}{98}$$

Reduce the fraction \frac{42}{98} to lowest terms by extracting and canceling out 14.

$$x=\frac{3}{7}$$

Steps for Completing the Square

Subtract 42x from both sides.

$$49x^{2}+9-42x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$49x^{2}-42x=-9$$

Divide both sides by 49.

$$\frac{49x^{2}-42x}{49}=-\frac{9}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\left(-\frac{42}{49}\right)x=-\frac{9}{49}$$

Reduce the fraction \frac{-42}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{6}{7}x=-\frac{9}{49}$$

Divide -\frac{6}{7}, the coefficient of the x term, by 2 to get -\frac{3}{7}. Then add the square of -\frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{7}x+\left(-\frac{3}{7}\right)^{2}=-\frac{9}{49}+\left(-\frac{3}{7}\right)^{2}$$

Square -\frac{3}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{7}x+\frac{9}{49}=\frac{-9+9}{49}$$

Add -\frac{9}{49} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{7}x+\frac{9}{49}=0$$

Factor x^{2}-\frac{6}{7}x+\frac{9}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{7}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{7}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{3}{7}=0 x-\frac{3}{7}=0$$

Add \frac{3}{7} to both sides of the equation.

$$x=\frac{3}{7} x=\frac{3}{7}$$

The equation is now solved. Solutions are the same.

$$x=\frac{3}{7}$$