Question

Solve x2 - 7x = -13.

x equals quantity of negative 7 plus or minus I square root of 6 all over 2

x equals quantity of negative 7 plus or minus I square root of 3 all over 2

x equals quantity of 7 plus or minus I square root of 6 all over 2

x equals quantity of 7 plus or minus I square root of 3 all over 2

Answer 1

@mathslover

Answer 2

@mathstudent55

Answer 3

The given equation can be better written as :

\(x^2 - 7x + \color{blue}{13} = -13 + \color{blue}{13} \\
x^2-7x+13 = \cancel{-13} + \cancel{\color{blue}{13}} \\
x^2 - 7x + 13 =0 \\ \)

Now, use quadratic equation to find the roots of the simplified equation.

\(\textbf{Roots} = \cfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Where the considered quadratic equation is : \(ax^2 + bx + c =0 \)

Compare the above equation (\(ax^2 + bx+c =0\) ) with the simplified equation (\(x^2-7x+13=0\)) and find : a, b and c and put the values in the quadratic equation.

Answer 4

-7 +- squareroot -7^2 - 4(1)(-13) / 2(1)

Answer 5

I will write it what you wrote above : (for better understanding)

\(\cfrac{-7 \pm \sqrt{(-7)^2 -4(1)(-13)}}{2(1)}\)

Answer 6

Now, see,

\(ax^2 + bx + c = 0\)
\(x^2 \space - 7x + 13 = 0\) or \(x^2 + (-7x) + 13 =0 \)

Compare the above two equations :

a = 1
b = -7
c = 13

Answer 7

You might have compared incorrectly. Rechec your process.

Answer 8

*Recheck

Answer 9

im really confused..

Answer 10

Where are you confused @kiaracakes120513 ?

Answer 11

i dont really know what i have to do @mathslover

Answer 12

You did well to put a, b and c in the equation .. but, you predicted the value of "b" incorrectly.

x^2 - 7x + 13 = 0

Here, b = - 7 not 7

Answer 13

okay

Answer 14

i did that

Answer 15

Now put a, b and c again in the quad. equation

Answer 16

\[-7 \pm \sqrt{-7^2 - 4(1)(-13)} \over 2(1)\]
@mathslover

Answer 17

See, it is -b in the numerator and we calculated b = - 7

so, -b becomes : -(-7) = 7

Answer 18

and what happens after that? im still a little confused but im sorta getting it..
@mathslover

Answer 19

It becomes :

\(\cfrac{7 \pm \sqrt{49 - 52} }{2} \)

=> \(\cfrac{7 \pm \sqrt{-3}}{2}\)

Answer 20

So, you get it as 7 plus minus square root of (-3) all over 2

Answer 21

ohh i get it now

Answer 22

can you help me with another please?

Answer 23

5x2 = -30x - 65

Answer 24

Transpose (-30x - 65) to LHS

5x^2 + 30x + 65 = 0

Take 5 common

Answer 25

so do i put that in the quadratic formula?

Answer 26

Yes! (After cancelling out from each term)

Answer 27

\[-30\pm \sqrt{30^2-4(5)(65)} \over 2(5)\]

Answer 28

is that right?

Answer 29

@mathslover

Answer 30

Yes, it is right Kiara.