Factor the following 2 quadratic equations:1) x2 + 6x – 16 = 02) 2x2 + 4x – 10 = 0 The second one is more difficult, right? That’s because the roots are not integers We need to solve the relation using the

Quadratic Formula

6.4 - The Quadratic Formula

2x2 + 4x – 10 = 0 2(x2 + 2x – 5) = 0 If we use (x + 1)2, we get x2 + 2x + 1 We need x2 + 2x – 5, so 6 less than above 2((x + 1)2 – 6) = 0 2(x + 1)2 – 12 = 0 2(x + 1)2 = 12 (x + 1)2 = 6 x + 1 = x = -1 Therefore the roots are x=-1+ & x =-1-

You can to solve using Vertex Form too:

Based on form ax2 + bx + c = 0 2x2 + 4x – 10 = 0 a = 2; b = 4; c = -10

Solve Using Quadratic Formula

Quadratic FormulaJust substitute the values for a, b and c into the formula, and

find the 2 values of x

Therefore, the roots of 2x2 + 4x – 10 = 0 are x = and x = x =

Solve Using Quadratic Formula

2x2 + 4x – 10 = 0a = 2; b = 4; c = -10

A rectangular field is going to be completely enclosed by 100m of fencing. Create a quadratic relation that shows how the area of the field will depend on its width. Then, determine the dimensions of the field that will results in an area of 575 m2. Round your answers to 2 decimal places.

Example #2

Let w be the width of the field If total perimeter is 100m, and

width is w, then length must be

Area = length x width = lw A = w(50 – w) A = 50w – w2

575 = 50w – w2

0 = -w2 + 50w – 575 Now use quadratic formula to find roots.

Example #2 cont’d

50 - w

w

0 = -w2 + 50w – 575 a = -1; b = 50; c = -575

Since 50 – w = 50 – 17.9 = 32.1, we can see that w = 17.9m and l = 32.1 m

Example #2 cont’d

The roots of a quadratic equation of the form ax2 + bx + c = 0 can be determined using the quadratic formula:

In Summary…

pg. 343 # 4acf, 5acf, 6, 8–10 (cd), 12-16, 18 - 20

6.4 Homework

Quadratic relations can have two, one or no x-intercepts.

6.5 Interpreting Quadratic Equation Roots

This graph shows the quadratic equation:-x2 + x + 6 = 0, and has 2 solutions: x = -

2 and x = 3

One/No Solutions/Root

This graph shows the quadratic equation:x2 – 6x + 9 = 0, and has 1 solution: x = 3

This graph shows the quadratic equation:

2x2 – 4x + 5 = 0, and has no solutions.

Challenge Question

How can you determine the number of solutions to a quadratic equation

without solving it?

The discriminant allows us to determine the number of real roots of each equation

◦ If D > 0, there are 2 real solutions◦ If D = 0, there is 1 real solution◦ If D < 0, there are no real solutions.

EX. 3x2 + 4x + 5 = 0

D < 0, so no solutions.

Discriminant

Determine the number of zeros for

Let’s try to solve by completing the square:

, but we need +17.5 +1.5 =

How can you tell the number to solutions by looking at the equation?

Example #2

𝑦=−2 (𝑥−4 )2−3Example #2 cont’d

Another way to solve: use Quadratic Formula.

(a = -2; b = 16; c = -35)

We cannot square root a negative number, so x does not exist. No solutions.

Example #2 cont’d

There are many ways to determine whether a quadratic equation has two, one or no real solutions, without solving the equation

A simple way is to use the determinant

◦ If D > 0, there are 2 real solutions◦ If D = 0, there is 1 real solution◦ If D < 0, there are no real solutions.

In Summary…