Exercise 126 Page 424

a We will use the Quadratic Formula to solve the given quadratic equation.

ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aWe first need to identify the values of a, b, and c. 2x^2-x-5=0 ⇕ 2x^2+( - 1)x+( - 5)=0 We see that a= 2, b= - 1, and c= - 5. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -1)±sqrt(( - 1)^2-4( 2)( - 5))/2( 2)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(2)(- 5))/2(2)

NegBaseToPosPow

(- a)^2 = a^2

x=1±sqrt(1-4(2)(- 5))/2(2)

Multiply

x=1±sqrt(1-8(- 5))/4

MultNegNegOnePar

- a(- b)=a* b

x=1±sqrt(1+40)/4

AddTerms

Add terms

x=1±sqrt(41)/4

We found that the solutions of the given equation are x_1= 1+sqrt(41)4 and x_2= 1-sqrt(41)4. Recall that any square root where the radicand is not a perfect square, such as 4, 9 or 16, will have infinite non-repeating decimals. Since 41 is not a perfect square, and we cannot rewrite the fractions, the solutions are irrational numbers.

b We want to solve the given equation for x. To do this we can start by using factoring. Then we will solve the equation using square roots.

Factoring

We want to factor the given equation. Let's start by gathering all terms on the left-hand side of the equation.

4x^2=4x-1 ⇔ 4x^2-4x+1=0 Factoring is much easier when our polynomial is a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to ask ourselves three questions.

Is the first term a perfect square?	4x^2= (2x)^2 ✓
Is the last term a perfect square?	1= 1^2 ✓
Is the middle term twice the product of 1 and 2x?	4x=2* 1* 2x ✓

As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is a subtraction sign in the middle. 4x^2-4x+1=0 ⇔ ( 2x- 1)^2=0

Solving

Since the equation is already written in perfect square form, we can now solve it using square roots.

(2x-1)^2=0

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((2x-1)^2)=sqrt(0)

CalcRoot

Calculate root

2x-1=0

AddEqn

LHS+1=RHS+1

2x=1

DivEqn

.LHS /2.=.RHS /2.

x=1/2

We found that x= 12. Recall that a rational number is a number that can be written as a fraction ab, where a and b are integers. Since this is the case, we can see that our solution is rational.

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Exercise 126 Page 424

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Factoring

Solving