Solve 1^3-6lambda^2+11lambda-6=0

$$1 ^ { 3 } - 6 \lambda ^ { 2 } + 11 \lambda - 6 = 0$$

Solve for λ

$$\lambda =1$$

$$\lambda =\frac{5}{6}\approx 0.833333333$$

Steps Using Factoring By Grouping

Calculate 1 to the power of 3 and get 1.

$$1-6\lambda ^{2}+11\lambda -6=0$$

Subtract 6 from 1 to get -5.

$$-5-6\lambda ^{2}+11\lambda =0$$

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

$$-6\lambda ^{2}+11\lambda -5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6\lambda ^{2}+a\lambda +b\lambda -5. To find a and b, set up a system to be solved.

$$a+b=11 ab=-6\left(-5\right)=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=6 b=5$$

Rewrite -6\lambda ^{2}+11\lambda -5 as \left(-6\lambda ^{2}+6\lambda \right)+\left(5\lambda -5\right).

$$\left(-6\lambda ^{2}+6\lambda \right)+\left(5\lambda -5\right)$$

Factor out 6\lambda in the first and -5 in the second group.

$$6\lambda \left(-\lambda +1\right)-5\left(-\lambda +1\right)$$

Factor out common term -\lambda +1 by using distributive property.

$$\left(-\lambda +1\right)\left(6\lambda -5\right)$$

To find equation solutions, solve -\lambda +1=0 and 6\lambda -5=0.

$$\lambda =1 \lambda =\frac{5}{6}$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 3 and get 1.

$$1-6\lambda ^{2}+11\lambda -6=0$$

Subtract 6 from 1 to get -5.

$$-5-6\lambda ^{2}+11\lambda =0$$

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.

$$-6\lambda ^{2}+11\lambda -5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 11 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$\lambda =\frac{-11±\sqrt{11^{2}-4\left(-6\right)\left(-5\right)}}{2\left(-6\right)}$$

Square 11.

$$\lambda =\frac{-11±\sqrt{121-4\left(-6\right)\left(-5\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$\lambda =\frac{-11±\sqrt{121+24\left(-5\right)}}{2\left(-6\right)}$$

Multiply 24 times -5.

$$\lambda =\frac{-11±\sqrt{121-120}}{2\left(-6\right)}$$

Add 121 to -120.

$$\lambda =\frac{-11±\sqrt{1}}{2\left(-6\right)}$$

Take the square root of 1.

$$\lambda =\frac{-11±1}{2\left(-6\right)}$$

Multiply 2 times -6.

$$\lambda =\frac{-11±1}{-12}$$

Now solve the equation \lambda =\frac{-11±1}{-12} when ± is plus. Add -11 to 1.

$$\lambda =-\frac{10}{-12}$$

Reduce the fraction \frac{-10}{-12} to lowest terms by extracting and canceling out 2.

$$\lambda =\frac{5}{6}$$

Now solve the equation \lambda =\frac{-11±1}{-12} when ± is minus. Subtract 1 from -11.

$$\lambda =-\frac{12}{-12}$$

Divide -12 by -12.

$$\lambda =1$$

The equation is now solved.

$$\lambda =\frac{5}{6} \lambda =1$$

Steps for Completing the Square

Calculate 1 to the power of 3 and get 1.

$$1-6\lambda ^{2}+11\lambda -6=0$$

Subtract 6 from 1 to get -5.

$$-5-6\lambda ^{2}+11\lambda =0$$

Add 5 to both sides. Anything plus zero gives itself.

$$-6\lambda ^{2}+11\lambda =5$$

Divide both sides by -6.

$$\frac{-6\lambda ^{2}+11\lambda }{-6}=\frac{5}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$\lambda ^{2}+\frac{11}{-6}\lambda =\frac{5}{-6}$$

Divide 11 by -6.

$$\lambda ^{2}-\frac{11}{6}\lambda =\frac{5}{-6}$$

Divide 5 by -6.

$$\lambda ^{2}-\frac{11}{6}\lambda =-\frac{5}{6}$$

Divide -\frac{11}{6}, the coefficient of the x term, by 2 to get -\frac{11}{12}. Then add the square of -\frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\lambda ^{2}-\frac{11}{6}\lambda +\left(-\frac{11}{12}\right)^{2}=-\frac{5}{6}+\left(-\frac{11}{12}\right)^{2}$$

Square -\frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$\lambda ^{2}-\frac{11}{6}\lambda +\frac{121}{144}=-\frac{5}{6}+\frac{121}{144}$$

Add -\frac{5}{6} to \frac{121}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\lambda ^{2}-\frac{11}{6}\lambda +\frac{121}{144}=\frac{1}{144}$$

Factor \lambda ^{2}-\frac{11}{6}\lambda +\frac{121}{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}.

$$\left(\lambda -\frac{11}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\lambda -\frac{11}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$\lambda -\frac{11}{12}=\frac{1}{12} \lambda -\frac{11}{12}=-\frac{1}{12}$$

Add \frac{11}{12} to both sides of the equation.

$$\lambda =1 \lambda =\frac{5}{6}$$