1. ## Solving a quadratic equation

Hi im having some difficulty in solving two quadratic equations shown below. i think i have answered the first one by using the quadratic formula

a) 2x²+15x+16=0 My Ans: -1.29 or -6.21

b) (2x-1)(3x+2) - (x+3)(2x+1) = 0 ?

Also a quick question, if a quadratic equation does not have a equal zero do i need to do something to the expression before trying to solve the quadratic.

Richard

2. 2x²+15x+16 = 0

$\displaystyle x = \frac {-15 \pm \sqrt{-15^2 - 4 * 2 * 16}}{2 * 2}$

$\displaystyle x = \frac {-15 \pm \sqrt{-15^2 - 4 * 2 * 16}}{4}$

$\displaystyle x = \frac {-15 \pm \sqrt{97}}{4}$

Using that it seems to be correct.

(b) (2x-1)(3x+2) - (x+3)(2x+1) = 0

Multiply the brackets ...

$\displaystyle (6x^2 + 4x - 3x - 2) - (2x^2 + x + 6x + 3) = 0$

$\displaystyle 6x^2 - 2x^2 + x - 7x - 2 - 3 = 0$

$\displaystyle 4x^2 - 6x - 5 = 0$

Using discriminant b^2 - 4ac .... 36 - 4 x 4 x - 5, it seems to have two roots.

Also a quick question, if a quadratic equation does not have a equal zero do i need to do something to the expression before trying to solve the quadratic.
Yes, re-arrange it to the left hand side.

3. thanks very much nice to know im going in the right direction
cheers
richard

4. ## A little Clarification

Originally Posted by ric115

Also a quick question, if a quadratic equation does not have a equal zero do i need to do something to the expression before trying to solve the quadratic.

In respone to your question about a quadrratic being equal to zero, the only thing that needs to be done if it does not equal zero is to move all of the terms to the left or right side of the equation and make it equal zero.

E.G. : x^2=-2x-1

Moving the terms yields

x^2+2x+1=0

now factoring......

(x+1)(x+1)=(x+1)^2=0

and this implies that

x=-1

5. Not really cos the equation doesnt have = sign in it. its shown as a expression shown below

The question asks me, to obtain the solution of the following quadratic equation

s² + 2abs + b²

This is the expression i was givin i have substituted 'a' and 'b' as they were mathamatical symbols.

Thanks

Richard

6. Well, that's the problem! There is a difference between a quadratic equation and a quadratic expression.

Equation: Something equal to something else

E.G. $\displaystyle x^2+2x+1=0$

Expression: A group of math mathematical terms

E.G. $\displaystyle x^2+4x-5$

Now, each one of these can be factored, but only one can be solved.

Understand?

7. so the answer would be that it cannot be solved as its not a quadratic equation? or would i need to factorise it?
Thanks
richard

8. Originally Posted by ric115
so the answer would be that it cannot be solved as its not a quadratic equation? or would i need to factorise it?
Thanks
richard