Hi.
Let us solve this quadratic
equation using
the quadratic formula.
Before that, we will write the
standard quadratic equation ax
square plus bx plus c equal 0.
The solution for this standard
quadratic form is given by the
quadratic formula x is equal
to negative b plus or minus
square root of b square
minus 4ac all over 2a.
I'm going to pan this and
move it aside, so we
can refer to that.
Now, our goal is to compare the
origin of the equation to
the standard form and try to
find values for a, b, and c.
So when you do that,
you find a is the
coefficient of x squared.
And in the given equation,
coefficient of x square is 2,
so a is same as 2.
Similarly, try to compare the
coefficient of x, and you get
b is negative 6.
And the constant term c
will be negative 3.
Now, we take these values and
try to plug them into the
quadratic formula to obtain the
solutions for the given
original equation.
Let us try to do that.
And I plug a, b, and c in the
quadratic formula, this is
what I get.
x will be negative b, negative
b would be negative of
negative 6 plus or minus square
root of b square is
squaring negative 6 minus 4
times a is 2 times c is
negative 3, all over 2a
will be 2 times 2.
We will now proceed to simplify
this further.
Negative of negative 6 would
be positive 6 plus or minus
square root.
Inside the square root we have
square of negative 6.
Square of negative 6 means
negative 6 times negative 6.
And so we will get 36.
Now, look at the next one.
Minus 4 times 2 times
negative 3.
So there are two negative
signs in the product.
And so negative times negative
will give us a positive.
And try to multiply the numbers
now, 4 times 2 is 8.
8 times 3 would be
24 all over 4.
And that would be 6 plus or
minus square root of, 36 plus
24 will give you
60, all over 4.
Let us try to simplify
root 60.
So when you take--
let me try to pan this--
OK, we're going to simplify
this further.
Right now we have 6 plus
or minus square root
of 60 all over 4.
You want to simplify root 60.
And in order to do that, we'll
have to factor 60.
We know 4 times 15 is 60,
and 15 can be further
factored as 3 times 5.
Now, this whole thing
is divided by 4.
And now we try to see
how further we can
simplify the radical.
3 and 5 are both prime
numbers, so
nothing can be done.
They have to stay inside
the 4 is a perfect
square, which is 2 square.
So when we split this radical
upon 4, 3, and 5--
and we can afford to
do that, because
these are all factors--
so square root of 4 will give
just the 2 outside.
And so we will have 6 plus or
minus 2 times square root of 3
times 5, which is
15, all over 4.
I'm going to pan this too.
And let's see how to simplify
the expression we got now, the
solution, which is 6 plus or
minus 2 root 15 all over 4.
Let's see how to simplify
this further.
Now, if you look at the
numerator, we have 6 and 2
root 15, which reads
as 2 times root 15.
Now, there's a 2 common
to both the terms.
And we can factor that out,
which will give 2 times 3 plus
or minus root 15 all over 4.
And now you can divide this 2
with the denominator, and get
2 in the denominator,
instead of 4.
So the solution set for the
given equation would be 3 plus
or minus root 15 divided by 2.
Write them withing
the set symbols.
So we have two solutions here.
3 plus root 15 divide by 2 and
3 minus root 15 divide by 2.
I hope this helps.
Thank you.
