Quadratic equations

[airbornewife29]

I need some help. Can anyone help me? We are doing quadratic equations. I am lost.

First problem; q(x)= 2x^2 + 3x - 1

The other problem;
Question:

To answer this question: Solve (x-1)^2=-49

Suppose your classmate wrote the first step of the solution as:

Take square root

x – 1 = 7 or x - 1 = -7

We now add 1 to each side of equations to solve for x.

x = 8 or x = -6

This is not correct since neither answer checks in the original equation. Can you correct this error? Please explain your thinking in simple language.

 

[Deleted member 4993]

I need some help. Can anyone help me? We are doing quadratic equations. I am lost.

First problem; q(x)= 2x^2 + 3x - 1

Can you write the solution of quadratic equation for us. If you cannot find it in your textbook - google it.

The other problem;
Question:

To answer this question: Solve (x-1)^2=-49

Suppose your classmate wrote the first step of the solution as:

Take square root << There is a negative sign in front of 49 ? what does that imply?

x – 1 = 7 or x - 1 = -7

We now add 1 to each side of equations to solve for x.

x = 8 or x = -6

This is not correct since neither answer checks in the original equation. Can you correct this error? Please explain your thinking in simple language.

 

[dbngshuroy]

I need some help. Can anyone help me? We are doing quadratic equations. I am lost.

First problem; q(x)= 2x^2 + 3x - 1

Problem1: q(x)= 2x^2 + 3x - 1 . I have to find the roots of the polynomial .Right?
Let's take a more general equation ax^2 + bx + c = 0 [a != 0]

Solving it we get

You can know it's proof from http://en.wikipedia.org/wiki/Quadratic_equation
your problem is a special case of the gen. equation. so putting a = 2 , b = 3 ,c = -1 you get the solution.

The other problem;
Question:

To answer this question: Solve (x-1)^2=-49

Suppose your classmate wrote the first step of the solution as:

Take square root

x – 1 = 7 or x - 1 = -7

We now add 1 to each side of equations to solve for x.

x = 8 or x = -6

This is not correct since neither answer checks in the original equation. Can you correct this error? Please explain your thinking in simple language.

Problem2:
(x-1)^2 = -49 does not imply x-1 = 7,-7 so You can not write the step.
let's do an different approach.
(x-1)^2 + 49 = 0
(x-1)^2 +7^2 = 0
2[(x-1)^2 +7^2] = 0

(x+6)^2 + (x-8)^2 = 0

Now here is 2 conditions.

• 1. x is Real
  2. x is not Real

if the first condition is true then
x+6 =0 and x-8 = 0 simultaneously which is not possible.so case 1 is false.

so here x is a complex number.
actually the solution is

(x-1)^2 = -49
Taking square root
x-1 = 7i,-7i
or x = 1 + 7i , 1- 7i

 

[BigGlenntheHeavy]

\(\displaystyle 1) \ q(x) \ = \ 2x^2+3x+1 \ = \ (2x+1)(x+1) \ \implies \ x \ = \ -1/2, \ x \ = \ -1.\)

\(\displaystyle 2) \ (x-1)^2 \ = \ -49, \ \implies \ (x-1) \ = \ \pm \ [-49]^{1/2}.\)

\(\displaystyle Hence, \ x-1 \ = \ \pm7i, \ \implies \ x \ = \ 1 \ \pm7i, \ i \ = \ imaginary.\)

\(\displaystyle Note \ for \ 2, \ no \ solution \ in \ real \ number \ land.\)