Solve 6(1-x^2)=5x

$$6 ( 1 - x ^ { 2 } ) = 5 x$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 6 by 1-x^{2}.

$$6-6x^{2}=5x$$

Subtract 5x from both sides.

$$6-6x^{2}-5x=0$$

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

$$-6x^{2}-5x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6\times 6=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=4 b=-9$$

Rewrite -6x^{2}-5x+6 as \left(-6x^{2}+4x\right)+\left(-9x+6\right).

$$\left(-6x^{2}+4x\right)+\left(-9x+6\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(-3x+2\right)+3\left(-3x+2\right)$$

Factor out common term -3x+2 by using distributive property.

$$\left(-3x+2\right)\left(2x+3\right)$$

To find equation solutions, solve -3x+2=0 and 2x+3=0.

$$x=\frac{2}{3} x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6 by 1-x^{2}.

$$6-6x^{2}=5x$$

Subtract 5x from both sides.

$$6-6x^{2}-5x=0$$

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.

$$-6x^{2}-5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+24\times 6}}{2\left(-6\right)}$$

Multiply 24 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\left(-6\right)}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\left(-6\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\left(-6\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{5±13}{-12}$$

Now solve the equation x=\frac{5±13}{-12} when ± is plus. Add 5 to 13.

$$x=\frac{18}{-12}$$

Reduce the fraction \frac{18}{-12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{5±13}{-12} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{-12}$$

Reduce the fraction \frac{-8}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{2}{3}$$

The equation is now solved.

$$x=-\frac{3}{2} x=\frac{2}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 6 by 1-x^{2}.

$$6-6x^{2}=5x$$

Subtract 5x from both sides.

$$6-6x^{2}-5x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-6x^{2}-5x=-6$$

Divide both sides by -6.

$$\frac{-6x^{2}-5x}{-6}=-\frac{6}{-6}$$

Dividing by -6 undoes the multiplication by -6.

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

Divide -5 by -6.

$$x^{2}+\frac{5}{6}x=-\frac{6}{-6}$$

Divide -6 by -6.

$$x^{2}+\frac{5}{6}x=1$$

Divide \frac{5}{6}, the coefficient of the x term, by 2 to get \frac{5}{12}. Then add the square of \frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{6}x+\left(\frac{5}{12}\right)^{2}=1+\left(\frac{5}{12}\right)^{2}$$

Square \frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{6}x+\frac{25}{144}=1+\frac{25}{144}$$

Add 1 to \frac{25}{144}.

$$x^{2}+\frac{5}{6}x+\frac{25}{144}=\frac{169}{144}$$

Factor x^{2}+\frac{5}{6}x+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{12}\right)^{2}=\frac{169}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Simplify.

$$x+\frac{5}{12}=\frac{13}{12} x+\frac{5}{12}=-\frac{13}{12}$$

Subtract \frac{5}{12} from both sides of the equation.

$$x=\frac{2}{3} x=-\frac{3}{2}$$