2021-12-26

To solve: $4{x}^{2}=\frac{15}{2}x+1$

Stuart Rountree

Step 1
To solve $4{x}^{2}=\frac{15}{2}x+1$
Multiply both sides by 2,
$2×4{x}^{2}=\left(\frac{15}{2}x+1\right)2$
$2×4{x}^{2}=8{x}^{2}$
By distributive property,
$\left(\frac{15}{2}x+1\right)2=2×\frac{15}{2}x+1×2$
$\left(\frac{15}{2}x+1\right)2=15x+2$
Replace $\left(\frac{15}{2}x+1\right)2$ with $15x+2$
Replace $2×4{x}^{2}$ with $8{x}^{2}$
$8{x}^{2}=15x+2$
Add $-15x-2$ on both sides,
$8{x}^{2}-15x-2=15x+2-15x-2$
$8{x}^{2}-15x-2=0$
Split the middle term
Replace
$8{x}^{2}-16x+x-2=0$
The above equation can be written as $8x×x-8x×2+1×x-1×2$
Apply distributive property for first two terms and last two terms,
$8x×x-8x×2+1×x-1×2=8x\left(x-2\right)+1\left(x-2\right)$
Again apply distributive property for $8x\left(x-2\right)+1\left(x-2\right)$
$8x\left(x-2\right)+1\left(x-2\right)=\left(8x+1\right)\left(x-2\right)$
So the factors are
$8{x}^{2}-15x-2=\left(8x+1\right)\left(x-2\right)$
To find the value of x,
Take $8x+1=0$
Subtract 1 on both sides,
$8x+1-1=0-1$
$8x=-1$
Divide by 8 on both sides,
$\frac{8x}{8}=\frac{-1}{8}$
$x=\frac{-1}{8}$
Take

Chanell Sanborn

Step 1
Given: $4{x}^{2}=\frac{15}{2}x+1$
Multiply both sides by 2
$4{x}^{2}×2=\frac{15}{2}x×2+1×2$
Simplify
$8{x}^{2}=15x+2$
Subtract 2 from both sides
$8{x}^{2}-2=15x+2-2$
Simplify
$8{x}^{2}-2=15x$
Subtract 15x from both sides
$8{x}^{2}-2-15x=15x-15x$
Simplify
$8{x}^{2}-15x-2=0$

$\sqrt{{\left(-15\right)}^{2}-4×8\left(-2\right)}=17$

Separate the solutions
${x}_{1}=\frac{-\left(-15\right)+17}{2×8}$
${x}_{2}=\frac{-\left(-15\right)-17}{2×8}$
$x=\frac{-\left(-15\right)+17}{2×8}:\phantom{\rule{1em}{0ex}}2$
$x=\frac{-\left(-15\right)-17}{2×8}:\phantom{\rule{1em}{0ex}}-\frac{1}{8}$
The solutions to the quadratic equation are:
Answer: $x=2,\phantom{\rule{1em}{0ex}}x=-\frac{1}{8}$

karton