week 6 – precalc 11 – Ruben's Blog

This week in math we learned how to solve a quadratic equation by completing the square. The example below demonstrated how to do so

$-2x^2+16x-3=0$

The first step in this equation is to factor out the -2 from the coefficients of x

$-2(x^{2}-8x)-3=0$

then take the sum number which is -8 and divide it by 2 and then square it

$\frac{-8}{2}x\frac{-8}{2}= 16$

then take 16 and add and subtract it to the original equation to create a zero pair

$-2(x^{2}-8x+16-16)-3=0$

then distribute the negative two too the negative sixteen in order for a factorable trinomial to remain inside the brackets

$-2(x^{2}-8x+16)32-3=0$

take everything outside of the brackets to the other side, and divide by negative to on both sides

$\frac{-2(x^2-8x+16)}{-2}=\frac{29}{-2}$ $x^2-8x+16=\frac{-29}{2}$

Then factor the remaining trinomial on the left side

$(x-4)^2=\frac{-29}{2}$

square root both sides in order get rid of the square

$\sqrt{(x-4)^2}=\frac{\sqrt{-29}}{\sqrt{2}}$

then on the right side multiply both the numerator and the denominator by square root 2 to get rid of the square root on the bottom

$x-4=\frac{\sqrt{-29}}{\sqrt{2}} x \frac{\sqrt{2}}{\sqrt{2}}$

$x-4=\frac{\sqrt{58}}{2}$

$x=+-\frac{\sqrt{58}}{2}+4$

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