Solve using the quadratic formula:x2 – 3x = 7x – 2?

I explain this exercise to you detailed, applying the general formula in this equation.

First that everything, we must organize the equation, this way:

x² - 3x = 7x - 2

x² - 3x - 7x + 2 = 0

x² - 10x + 2 = 0

Now this formula:

....-b ± √b²-4ac

x=------------------

.............2a

Where the values are the following ones, comparing with the quadratic form: ax² + bx+c.

a=1

b =-10

c=2

Now we replace the values in the formula and resolve:

....-(-10) ± √(-10)²-4(1)(2)

x=---------------------------------

...................2(1)

We multiply what is in of the parenthesis, this way:

....10 ± √100-8

x=-----------------

................2

We remain and then we realize that it does not have root square the number ninety two, therefore, is indicated indicated and is continued by the procedure.

Now both values are looked independently, of following way:

X1 = (10+√92)/5

x2 = (10-√92)/5

Therefore this they are the values of this quadratic equation.

I hope that my explanations use you as usefulness, greet from Colombia! That has a happy day and good bye ^^.

Question Number 1 :

For this equation x^2 - 3*x = 7*x - 2 , answer the following questions :

A. Find the roots using Quadratic Formula !

Answer Number 1 :

First, we have to turn equation : x^2 - 3*x = 7*x - 2 , into a*x^2+b*x+c=0 form.

x^2 - 3*x = 7*x - 2 , move everything in the right hand side, to the left hand side of the equation

<=> x^2 - 3*x - ( 7*x - 2 ) = 0 , which is the same with

<=> x^2 - 3*x + ( - 7*x + 2 ) =0 , now open the bracket and we get

<=> x^2 - 10*x + 2 = 0

The equation x^2 - 10*x + 2 = 0 is already in a*x^2+b*x+c=0 form.

In that form, we can easily derive that the value of a = 1, b = -10, c = 2.

1A. Find the roots using Quadratic Formula !

Remember the formula,

x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)

Since a = 1, b = -10 and c = 2,

we need to subtitute a,b,c in the abc formula, with thos values.

So x1 = (-(-10) + sqrt( (-10)^2 - 4 * (1)*(2)))/(2*1) and x2 = (-(-10) - sqrt( (-10)^2 - 4 * (1)*(2)))/(2*1)

Which can be turned into x1 = ( 10 + sqrt( 100-8))/(2) and x2 = ( 10 - sqrt( 100-8))/(2)

Which is the same with x1 = ( 10 + sqrt( 92))/(2) and x2 = ( 10 - sqrt( 92))/(2)

Which is the same with x1 = 5 + sqrt(26) and x2 = 5 - sqrt(26)

So we get x1 = ( 10 + 9.59166304662544 )/(2) and x2 = ( 10 - 9.59166304662544 )/(2)

The answers are x1 = 9.79583152331272 and x2 = 0.204168476687281

ax^2+bx+c=0 as a consequence a=a million b=-7 c=5 (x-A)(x-B)=x^2-Bx-Ax+AB=x^2-x(A+B)+AB so which you're finding for 2 numbers such that A+B=7 and A*B=5 yet by skill of the rational root theorem, the candidate roots are 5,a million,-5,-a million try some mixtures: 5+a million=6 and 5*a million=5 5-a million=4 and 5(-a million)=-5 -5+a million=-4 and -5(a million)=-5 -5-a million=-6 and (-5)(-a million)=5 so not one of the rational root possibilities artwork. the answer is irrational and should be solved by skill of the quadratic equation that's simlply polishing off the sq. with variables. to end the sq.: (x-w)^2=x^2+-2xw+w^2 upload and subtract w^2 by skill of finding at like words, -2w=-7 so w=7/2 and w^2=40 9/4 so the equtoin will become: x^2-7x+40 9/4-40 9/4+5=0 (x-7/2)^2-40 9/4+20/4=0 (x-7/2)^2=29/4 x-7/2=+/-sqrt(29)/2 x=(7 +/- sqrt(29))/2

x^2 - 3x = 7x - 2

x^2 - 3x - 7x + 2 = 0

x^2 - 10x + 2 = 0

x = [-b ±√(b^2 - 4ac)]/2a

a = 1

b = -10

c = 2

x = [10 ±√(100 - 8)]/2

x = [10 ±√92]/2

x = [10 ±9.59]/2 (approx.)

x = [10 + 9.59]/2

x = 19.59/2

x = 9.795

x = [10 - 9.59]/2

x = 0.41/2

x = 0.205

∴ x = 0.205 , 9.795

x^2-10x+2=0

x=10+-sqrt(100-8)/2

= 10+ -sqrt(92)/2