Thread: Need step-by-step solution

Solving for roots. I need to explain the procedure to my twelve year old bro who's failing his math class, unfortunately I haven't been able to be of much help, so I'm hoping that you guys can. plus the kind of roots the last three equations have. Thanks

a) 6(80-30) = 48-x

b) 2x^2-13x+6 = 0; by factoring

c) 3x^2 - 24x + 48 = 0

d) 4x^2 + 7 = 0

e) 4 (7x-4)^2 +1 = 9

Originally Posted by archistrategos214

Solving for roots. I need to explain the procedure to my twelve year old bro who's failing his math class, unfortunately I haven't been able to be of much help, so I'm hoping that you guys can. plus the kind of roots the last three equations have. Thanks

a) 6(80-30) = 48-x

Is there something missing here, this is a linear (1st order problem while
the rest are quadratic (2nd order) problems?

Any way this goes something like:

1. Evaluate the left hand side:

6(80-30) = 6*50 = 300 = 48-x

so we now have:

300 = 48-x

If we subtract 48 from both sides of this equation we still have equality
so:

300 - 48 = 48 - x -48

or:

252 = -x.

Now multiply through by -1 to get:

(-1)*252 = -252 = (-1)(-x) = x,

or:

x = -252.

RonL

Originally Posted by archistrategos214

b) 2x^2-13x+6 = 0; by factoring

We wish to factor this into the form:

(ax+b)(cx+d) = 2x^2-13x+6 = 0

Now if this is to be solved by factoring it means that we are looking
for integers for a, b, c and d.

So d and d are two different factors of 6 as when we multiply out the
brackets on the far left the term independent of x is bd (but with both having
either a + or - sign in front of it, the same sign for both). The factors of 6
are 1, 2, 3, 6.

Similarly a and c are two different of 2 as when we multiply out the brackets
the coefficient of x^2 is ac. The factors of 2 are 1, 2.

Now trying out all the possibilities we find the one that gives us -13 for the
coefficient of the x term. The result of this is:

(2x - 1)(x -6) = 2x^2-13x+6 = 0.

Now if a product is zero one or other of the terms must be zeros, so
we have either

(2x - 1) = 0,

in which case x = 1/2,

or

(x-6) = 0,

when x = 6.

So the roots of:

2x^2-13x+6 = 0

are x = 1/2 and x = 6.

RonL

Originally Posted by archistrategos214

c) 3x^2 - 24x + 48 = 0

Divide both sides by 3.
That means each summand will get divided by three:
x^2-8x+16=0
Look at the factors of 16:
(1,16),(2,8),(4,4)
We see that using 4,4 when subtracting gives -8 thus,
(x-4)(x-4)=0
Make each factor equal to zero:
x-4=0 and x-4=0
Thus,
x=4

This means any possible solution must be 4 but not necessarily the other way around. We check and indeed see that 4 is a solution to this equation.

Originally Posted by archistrategos214

d) 4x^2 + 7 = 0

Subtract 7 thus,
4x^2=-7
Divide by 4 thus,
x^2=-7/4
Take square root of both sides,
x=+\- sqrt(-7/4)=+\- i*sqrt(7)/2

e) 4 (7x-4)^2 +1 = 9

Subtract 1,
4(7x-4)^2=8
Divide both sides by 4,
(7x-4)^2=2
Square root,
(7x-4)=+\- sqrt(2)
Add 4,
7x=4+\- sqrt(2)
Divide by 7,
x=(4/7)+\- sqrt(2)/7