Solving Quadratic Equations

a

Let's start by dividing both sides by $3$ to get the $x^{2}$ -term alone. We must remember that we then get two solutions.

3 x^{2} = 75

LHS / 3 = RHS / 3

x^{2} = 25

LHS = RHS

x = \pm 25

CalcRootCalculate root

x = \pm 5

The solutions are then

x = 5

and

x = - 5 .

b

First, we will add

16

to both sides. After that we will divide both sides by

4

to get the

x^{2}

-term alone.

4 x^{2} - 16 = 0

LHS + 16 = RHS + 16

4 x^{2} = 16

LHS / 4 = RHS / 4

x^{2} = 4

LHS = RHS

x = \pm 4

CalcRootCalculate root

x = \pm 2

c

We will use inverse operations to get the $x^{2}$ -term alone. First we must undo the subtraction and then the division.

\frac{x ^{2}}{6} - 4 = 2

LHS + 4 = RHS + 4

\frac{x ^{2}}{6} = 6

LHS \cdot 6 = RHS \cdot 6

x^{2} = 36

LHS = RHS

x = \pm 36

CalcRootCalculate root

x = \pm 6

d

To solve the equation we must use inverse operations to get all the

x^{2}

-terms onto the same side.

x^{2} = 15 - 2 x^{2}

LHS + 2 x^{2} = RHS + 2 x^{2}

3 x^{2} = 15

LHS / 3 = RHS / 3

x^{2} = 5

LHS = RHS

x = \pm 5

Solving Quadratic Equations 1.2 - Solution