Solve b^2-6b+8 | Microsoft Math Solver (2024)

p+q=-6 pq=1\times 8=8

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb+8. To find p and q, set up a system to be solved.

-1,-8 -2,-4

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 8.

-1-8=-9 -2-4=-6

Calculate the sum for each pair.

p=-4 q=-2

The solution is the pair that gives sum -6.

\left(b^{2}-4b\right)+\left(-2b+8\right)

Rewrite b^{2}-6b+8 as \left(b^{2}-4b\right)+\left(-2b+8\right).

b\left(b-4\right)-2\left(b-4\right)

Factor out b in the first and -2 in the second group.

\left(b-4\right)\left(b-2\right)

Factor out common term b-4 by using distributive property.

b^{2}-6b+8=0

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.

b=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}

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.

b=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}

Square -6.

b=\frac{-\left(-6\right)±\sqrt{36-32}}{2}

Multiply -4 times 8.

b=\frac{-\left(-6\right)±\sqrt{4}}{2}

Add 36 to -32.

b=\frac{-\left(-6\right)±2}{2}

Take the square root of 4.

b=\frac{6±2}{2}

The opposite of -6 is 6.

b=\frac{8}{2}

Now solve the equation b=\frac{6±2}{2} when ± is plus. Add 6 to 2.

b=4

Divide 8 by 2.

b=\frac{4}{2}

Now solve the equation b=\frac{6±2}{2} when ± is minus. Subtract 2 from 6.

b=2

Divide 4 by 2.

b^{2}-6b+8=\left(b-4\right)\left(b-2\right)

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

x ^ 2 -6x +8 = 0

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.

r + s = 6 rs = 8

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 = 3 - u s = 3 + u

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

(3 - u) (3 + u) = 8

To solve for unknown quantity u, substitute these in the product equation rs = 8

9 - u^2 = 8

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

-u^2 = 8-9 = -1

Simplify the expression by subtracting 9 on both sides

u^2 = 1 u = \pm\sqrt{1} = \pm 1

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

r =3 - 1 = 2 s = 3 + 1 = 4

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