Mastering Solving Simple Quadratic Equations with Mathleaks

Equation	Rewrite as $x^{2} = d$	Value of $d$	Number of Real Solutions
$- 4 x^{2} + 5 = 5$	$x^{2} = 0$	$d = 0$	One
$5 x^{2} - 125 = 0$	$x^{2} = 25$	$d = 25 \Rightarrow d > 0$	Two
$- 3 x^{2} - 27 = 0$	$x^{2} = - 9$	$d = - 9 \Rightarrow d < 0$	Zero

Equation

Rewrite as

x^{2} = d

Value of

d

Number of Real Solutions

- 4 x^{2} + 5 = 5

x^{2} = 0

d = 0

One

5 x^{2} - 125 = 0

x^{2} = 25

d = 25 \Rightarrow d > 0

Two

- 3 x^{2} - 27 = 0

x^{2} = - 9

d = - 9 \Rightarrow d < 0

Zero

Equation	Rewrite
$- x^{2} + 4 = 0$	$- 1 (x - 0)^{2} + 4 = 0$
$3 x^{2} - 6 x + 5 = 0$	$3 (x - 1)^{2} + 2 = 0$
$5 x^{2} = 3 x + 1$	$5 (x - \frac{3}{1 0})^{2} + (- \frac{2 9}{2 0}) = 0$
$4 x^{2} + 4 x = - 2$	$4 (x - (- \frac{1}{2}))^{2} + 1 = 0$
$4 x^{2} + 4 x = - 2$	$4 (x - (- \frac{1}{2}))^{2} + 1 = 0$
$10 x^{2} + 160 = 80 x$	$10 (x - 4)^{2} + 0 = 0$

Equation

Rewrite

- x^{2} + 4 = 0

- 1 (x - 0)^{2} + 4 = 0

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

3 (x - 1)^{2} + 2 = 0

5 x^{2} = 3 x + 1

5 (x - \frac{3}{1 0})^{2} + (- \frac{2 9}{2 0}) = 0

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

4 (x - (- \frac{1}{2}))^{2} + 1 = 0

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

4 (x - (- \frac{1}{2}))^{2} + 1 = 0

10 x^{2} + 160 = 80 x

10 (x - 4)^{2} + 0 = 0

Consider the following diagram.

Concentric squares with outer side length of 14

The area of the inner square is

25 %

the area of the bigger square. Find the side length

ℓ

of the inner square.

Let's start by finding the area of the bigger square A_b. To do so, we will use the formula for the area of a square.

Let A_i represent the area of the inner square. This area is 25 % the area of the bigger square A_b.

The area of the inner square is 49 square meters. Since l is the side length of this square, we can write an equation showing the relationship between l and 49. l^2=49 We will solve this equation by taking square roots of both sides of the equation.

The solutions to the equation are l=7 and l=- 7. However, a negative value does not make sense in this context, since a measurement of length cannot be negative. Therefore, the side length of the inner square is 7 meters.

Consider the following diagram.

Right triangle with legs equal to 4x and 3x and hypotenuse 40

Find the value of

x .

The triangle in the given diagram is a right triangle, so we can use the Pythagorean Theorem to write a relationship for its side lengths. a^2+b^2=c^2 Here, a and b are the legs of the triangle and c its hypotenuse. Let's substitute the values from the diagram into the above formula and simplify.

The resulting is a quadratic equation. Let's solve it by taking square roots on both sides.

Since length measurements must be positive, the negative solution does not make sense in this scenario. Therefore, the answer is x=8. Incidentally, this means that the leg lengths of the triangle are 4*8=32 units and 3*8=24 units.

Consider the following diagram.

The area of the circle is

30

square inches. Find its radius

r .

Round the answer to two decimal places.

Let's begin by recalling the formula for the area of a circle. A=π r^2 We are told that the area of this particular circle is 30 square inches. Let's substitute this into the formula to write a partial equation for the area of this circle. A=π r^2 ⇒ 30=π r^2 The result is a quadratic equation that can be solved by taking square roots of both sides. First we need to isolate the radius r on one side of the equation. Let's do it!

Since a radius cannot be negative, a negative solution does not make sense in this scenario. Therefore, the radius of the circle is about 3.09 inches.

Solving Quadratic Equations Using Square Roots

Catch-Up and Review

Determining the Number of Solutions

Simple Quadratic Equations

Number of Solutions of a Simple Quadratic Equation

Proof

$d > 0$

$d = 0$

$d < 0$

Number of Solutions

Hint

Solution

How Many Solutions?

Solving Quadratic Equations with Square Roots

Rational Solutions

Hint

Solution

Irrational Solutions

Hint

Solution

Solving Simple Quadratic Equations

Solving Other Type of Quadratic Equation

Hint

Solution

A Process Called Completing the Square

Solving Quadratic Equations Using Square Roots

Recommended exercises

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

Solving Quadratic Equations Using Square Roots

Catch-Up and Review

Proof

d>0

d=0

d<0

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Solving Quadratic Equations Using Square Roots

Recommended exercises

$d > 0$

$d = 0$

$d < 0$