Exercise 88 Page 447

Given the solutions from the Quadratic Formula, we should not use the zero product property as the zeroes are not integers. Instead we will complete the square. To complete the square, we equate

y

with

0

and then rewrite it in the following format.

x^{2} + b x = c

Let's start by doing this.

y = 5 x^{2} + 7 x - 6

Substitute

$y = 0$

0 = 5 x^{2} + 7 x - 6

RearrangeEqn

Rearrange equation

5 x^{2} + 7 x - 6 = 0

AddEqn

$LHS + 6 = RHS + 6$

5 x^{2} + 7 x = 6

DivEqn

$LHS / 5 = RHS / 5$

x^{2} + 1.4 x = 1.2

Next we will rewrite the left-hand side so that it can be written as a square. To do that, we have to add the square of half the coefficient to

x,

which is

(\frac{1 . 4}{2})^{2} = 0 . 7^{2},

to both sides of the equation. This is called completing the square.

x^{2} + 1.4 x = 1.2

AddEqn

$LHS + 0 . 7^{2} = RHS + 0 . 7^{2}$

x^{2} + 1.4 x + 0 . 7^{2} = 1.2 + 0 . 7^{2}

▼

Solve for

x

SplitIntoFactors

Split into factors

x^{2} + 2 (0.7) x + (0.7)^{2} = 1.2 + (0.7)^{2}

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

(x + 0.7)^{2} = 1.2 + (0.7)^{2}

CalcPow

Calculate power

(x + 0.7)^{2} = 1.2 + 0.49

AddTerms

Add terms

(x + 0.7)^{2} = 1.69

SqrtEqn

$LHS = RHS$

x + 0.7 = \pm 1.3

SubEqn

$LHS - 0.7 = RHS - 0.7$

x = \pm 1.3 - 0.7

StateSol

State solutions

x_{1} = - 2 x_{2} = 0.6

Both methods gives the same answer.