Unlocking Solving Quadratic Equations Using the Quadratic Formula

Here are a few recommended readings before getting started with this lesson.

Square Root
Quadratic Equation
Standard Form of a Quadratic Function
Completing the Square

Try a few practice exercises as a warm-up!

4^{2} - 4 (1) (2)

by using the properties of square roots to remove the perfect-square factor.

b Consider the quadratic function

y = (x + 2)^{2} - 1 .

Of the following expressions, which represents the same function written in standard form?

c Identify the coefficient

b

of the quadratic equation

x - 5 = 2 (x - 3)^{2}

when written in standard form

a x^{2} + b x + c = 0 .

$x = \frac{- 3 2 \pm 1 6}{- 4}$
$x = \frac{- 3 2 + 1 6}{- 4}$	$x = \frac{- 3 2 - 1 6}{- 4}$
$x = \frac{- 1 6}{- 4}$	$x = \frac{- 4 8}{- 4}$
$x = 4$	$x = 12$

x = \frac{- 3 2 \pm 1 6}{- 4}

x = \frac{- 3 2 + 1 6}{- 4}

x = \frac{- 3 2 - 1 6}{- 4}

x = \frac{- 1 6}{- 4}

x = \frac{- 4 8}{- 4}

x = 4

x = 12

Value of the Discriminant	Number of Real Solutions
$b^{2} - 4 a c > 0$	$2$
$b^{2} - 4 a c = 0$	$1$
$b^{2} - 4 a c < 0$	$0$

Value of the Discriminant

Number of Real Solutions

b^{2} - 4 a c > 0

2

b^{2} - 4 a c = 0

1

b^{2} - 4 a c < 0

0

The function $p (x) = x^{2} - 2 x - 15$ has been graphed below along with a straight line.

a parabola and a straight line intercepting the vertex and positive x-intercept of the line

The straight line passes through the vertex and one of the

x -

intercepts of the quadratic function. Write the equation of the line in slope-intercept form.

To determine the equation of a line, we need to know at least two points that lie on the line. We are told that the line goes through the vertex and one of the zeros of the parabola p(x).

Zeros of the Parabola

To find the zeros, we will set p(x) equal to 0 and solve for x using the Quadratic Formula.

The quadratic function intercepts the x-axis at x_1=5 and x_2=-3.

Vertex of the Parabola

The x-coordinate of the vertex of the parabola is given by the equation of the axis of symmetry of the parabola. We can determine its equation by using the formula x_s = x_1+x_22, where x_1 and x_2 are the already found zeros.

Therefore, x=1 is the x-coordinate of the vertex. We can determine the y-coordinate of the vertex by substituting x=1 into the function rule and evaluating.

The vertex of the function is (1,-16).

Equation of the Line

We are ready to determine the slope of the line. We can see in the diagram that the line intersects the x-axis at the positive zero of the parabola. Therefore, let's substitute the coordinates of the x-intercept (5,0) and the vertex (1,-16) into the slope formula to find the slope of the line.

Now that we know the line's slope, we can determine its y-intercept by substituting it and one of the points into the slope-intercept form of an equation and solving for b.

Finally, we can write the complete equation. y = 4x-20

Consider the quadratic equation.

x^{2} + k x + 9 = 0

For what values of

k

is there one real solution to the equation?

For what values of

k

are there zero real solutions to the equation? Write the answer as a compound inequality.

The number of solutions to a quadratic equation can be determined by examining the discriminant, which is included in the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a The value of the discriminant will help us determine the number of real solutions to a quadratic equation.

Value of the Discriminant	Number of Real Solutions
b^2-4ac > 0	Two
b^2-4ac = 0	One
b^2-4ac < 0	Zero

Let's determine the values of a, b, and c in the standard form of the given quadratic equation, which we will then substitute into the formula for the discriminant. x^2 + kx + 9 = 0 ⇓ 1x^2 + kx + 9 = 0 Now we will substitute a= 1, b= k, and c= 9 into the formula for the discriminant and evaluate.

The discriminant can be described by the expression k^2-36. If we set it equal to zero and solve for k, we can determine the values of k for which the equation has only one real solution.

When k=-6 or k=6, the quadratic equation has one real solution.

In order for the equation to have no real solutions, the discriminant must be negative. This allows us to write the following inequality. k^2-36 < 0 Let's now solve for k.

Now we have a quadratic inequality, which can sometimes be difficult to solve algebraically. It is usually easier to solve these inequalities graphically. If we draw y=k^2 and y=36, we see that they intersect when k=±6.

Now we can see that y=k^2 is less than 36 when k>-6 and k<6. We can write this as the following compound inequality. -6

Consider the following graph of a quadratic function.

Which equation(s) could describe the function of the graph?

A . B . C . D . E . F . f (x) = x^{2} - 4 x + 6 f (x) = - x^{2} - 4 x + 6 f (x) = x^{2} - 6 x + 6 f (x) = x^{2} - 10 x - 6 f (x) = x^{2} + 5 x + 8 f (x) = - x^{2} + 2 x + 4

Let's first list everything we know about the graph of the quadratic function.

The vertex of the function is a minimum.
The graph intercepts the positive part of the y-axis.
The graph has no x-intercepts.
The x-coordinate of the vertex is positive.

We can use this information to exclude the potential functions one at the time. Notice that we have not been given any points the parabola passes through. We only need to identify which functions match our criteria.

Minimum Value

A parabola with a minimum value opens upwards has a positive leading coefficient. Since our parabola opens upwards, this means that we can exclude B and F, where the leading coefficient is negative. & A. f(x)=x^2-4x+6 & B. f(x)= - x^2-4x+6 & * & C. f(x)=x^2-6x+6 & D. f(x)=x^2-10x-6 & E. f(x)=x^2+5x+8 & F. f(x)= - x^2+2x+4 & *

Positive y-intercept

Since the curve intercepts the y-axis on its positive side, we know that the function must have a positive constant term. Therefore, we can exclude option D where the constant is negative. & A. f(x)=x^2-4x+6 & C. f(x)=x^2-6x+6 & D. f(x)=x^2-10x - 6 & * & E. f(x)=x^2+5x+8

No x-intercepts

To determine which of the remaining functions have no x-intercepts, we will have a look at the discriminant of the functions. First we set f(x) equal to zero and identify the values of a, b, and c.

Function	a	b	c
A. 1x^2 - 4x+ 6=0	1	-4	6
C. 1x^2 - 6x+ 6=0	1	-6	6
E. 1x^2 + 5x+ 8=0	1	5	8

The function has no x-intercepts if the value of the discriminant is less than zero. Let's calculate the discriminate for each of the three remaining functions. Recall that the formula for the discriminant of a function is b^2-4ac. If the discriminant is greater or equal to zero, we know that the function does not model the given graph. A. & ( -4)^2-4( 1)( 6)=-8 C. & ( -6)^2-4( 1)( 6)=10 * E. & 5^2-4( 1)( 8)=-7 As we can see, the function in option C has two x-intercepts because the discriminant is greater than 0. Therefore, we can exclude this option as well. & A. f(x)=x^2-4x+6 && & E. f(x)=x^2+5x+8 &&

Coordinates of the Vertex

The x-coordinate of the vertex is given by the equation of the axis of symmetry. It can be calculated by using the following formula. x_s=- b/2a We have already identified the values of a and b for both A and E. If we substitute these values into the formula, we can determine their axes of symmetry. & A. x_s=- ( -4)/2( 1)=2 [1em] & E. x_s=- 5/2( 1)=-2.5 Now we can exclude option E as well, since its axis of symmetry is located on the negative side of the x-axis. & A. f(x)=x^2-4x+6 && & E. f(x)=x^2+5x+8 && * This leaves only option A. To confirm that the function in option A does indeed meet all of our conditions, we need to check if the y-coordinate of the vertex is positive. Therefore, let's substitute the axis of symmetry x=2 into the function rule and solve for f(x), the y-value of the vertex.

Similar to the function in the graph, the y-coordinate of the vertex is positive. This means that the only function that could represent the graph is given in option A. & A. f(x)=x^2-4x+6 && ✓ & B. f(x)=- x^2-4x+6 && * & C. f(x)=x^2-6x+6 && * & D. f(x)=x^2-10x-6 && * & E. f(x)=x^2+5x+8 && * & F. f(x)=- x^2+2x+4 && *

Solve the following equation for

x

when

x^{2} = t .

x^{4} - 5 x^{2} + 4 = 0

Let's perform the variable substitution x^2=t. Before we can do that, we must rewrite x^4 in terms of x^2.

Now we have a corresponding quadratic equation that we can solve by using the Quadratic Formula.

We found that the solutions to the quadratic equation are t=4 and t=1. We now need to substitute these solutions into the substitution equation x^2=t and solve for x. That way, we will find the solutions to the original equation. x^2=1 ⇒ x=±1 x^2=4 ⇒ x=±2 We have a total of four solutions to the original equation: x=-1, x=1, x=-2, and x=2.

The given equation can also be solved by rewriting the x^2-term and factoring. Start by writing -5x^2 as -4x^2 - x^2. x^4-5x^2+4=0 ⇕ x^4 - 4x^2 - x^2 + 4 = 0 Next, notice that x^2 can be factored out from the first two terms. Also, factor out -1 from the next two terms. x^4 - 4x^2 - x^2 + 4 = 0 ⇕ x^2( x^2 - 4 ) - (x^2 - 4 ) = 0 Now, ( x^2 - 4 ) can be factored out. Then, use the difference of squares formula.

Finally, the Zero Product Property can be applied to write all four solutions to the original equation. lx+2 = 0 x-2 = 0 x+1 = 0 x-1 = 0 ⇔ lx = -2 x = 2 x = -1 x = 1

Solving Quadratic Equations With the Quadratic Formula

Catch-Up and Review

Is There a Solution?

The Quadratic Formula

Quadratic Formula

Proof

Solving a Quadratic Equation Using the Quadratic Formula

Hint

Solution

Solving a Quadratic Equation Not in Standard Form

Hint

Solution

Solving a Quadratic Equation Using the Quadratic Formula

Discriminant of a Quadratic Equation

Discriminant

Determining Whether There is a Solution Without Solving

Hint

Solution

Determining the Number of Real Solutions of a Quadratic Equation

Determining the Number of Solutions of a Quadratic Equation Without Solving

Hint

Solution

Zeros of the Parabola

Vertex of the Parabola

Equation of the Line

Minimum Value

Positive y-intercept

No x-intercepts

Coordinates of the Vertex

Solving Quadratic Equations With the Quadratic Formula

Recommended exercises

	10 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions