Hey i am attempting to CTS 4x^{2} +24x-18=0
i divide by four to make the a coefficient zero but then get lost at the 18/4 and dont know what do with that.
Any help is appreciated
Foe such questions adopt the folloing algorithm.
Step - 1: Make coefficient = 1. in this case : divide by 4: x^2 + 6x - 9/2 = 0
step - 2: transfer the constant term to the RHS: we will get x^2 + 6x = 9/2
Step 3: Add the square of half the coefficient of x on both the sides of the equation. in this case the coefficient of x = 6; Half the coefficient of x = 3 and square of this is 9. Thus we have: x^2 + 6x + 9= 9/2 + 9
That gives ( x + 3 )^2 = 27/2
For this particular problem here is a possible approach.
$\displaystyle 4x^2 + 24x = 18 $
fnd $\displaystyle \sqrt{4} = 2 $ , now divide 24 by $\displaystyle (2 \sqrt{4}) = 4 $ , $\displaystyle \frac{24}{4} = 6 $ , then square 6, 6^2 = 36
now look where sqrt(4) = 2 , 6 and 36 go...
$\displaystyle (2x + 6)^2 - 36 = 18 $
$\displaystyle (2x + 6)^2 = 54 $
The square has been completed... extract the roots...
$\displaystyle 2x + 6 = \pm \sqrt{54} $
$\displaystyle 2x = -6 \pm 3 \sqrt{6} $
$\displaystyle x = \frac{-6 \pm 3 \sqrt{6}}{2} $
Using the standard method...
$\displaystyle \displaystyle \begin{align*} 4x^2 + 24x - 18 &= 0 \\ 4 \left( x^2 + 6x - \frac{9}{2} \right) &= 0 \\ 4 \left( x^2 + 6x + 3^2 - 3^2 - \frac{9}{2} \right) &= 0 \\ 4 \left[ \left( x + 3 \right) ^2 - 9 - \frac{9}{2} \right] &= 0 \\ 4 \left[ \left( x + 3 \right) ^2 - \frac{27}{2} \right] &= 0 \\ 4 \left( x + 3 \right) ^2 - 54 &= 0 \end{align*}$
Go from here...