Question

Find the solution set for 8x^-9x=-1

Answer 1

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{8},\:x=1$

Step-by-step explanation:

Given the equation

$8x^2-9x=-1\:\:$

Add 1 to both sides

$8x^2-9x+1=-1+1$

$8x^2-9x+1=0$

as

$\left(8x-1\right)\left(x-1\right)=8x^2-9x+1$

so the equation becomes

$\left(8x-1\right)\left(x-1\right)=0$

Using the zero factor principle:

$\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$8x-1=0\quad \mathrm{or}\quad \:x-1=0$

solving

$8x-1=0$

$8x=1$

$\frac{8x}{8}=\frac{1}{8}$

$x=\frac{1}{8}$

also solving

$x-1=0$

$x=1$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{8},\:x=1$