Monique Slaughter

2021-12-29

To solve:
$x\left(5x+2\right)=3$

Alex Sheppard

Step 1
Given: $x\left(5x+2\right)=3$
By distributive property,
$x\left(5x+2\right)=x×5x+x×2$
$x\left(5x+2\right)=5{x}^{2}+2x$
Replace $x\left(5x+2\right)$ with $5{x}^{2}+2x$
To solve $5{x}^{2}+2x=3$
Subtract 3 on both sides,
$5{x}^{2}+2x-3=3-3$
$5{x}^{2}+2x-3=0$
To solve, $5{x}^{2}+2x-3=0$
Split the middle term
Replace
$5{x}^{2}+5x-3x-3=0$
The above equation can be written as $5x×x+1×5x-3×x-3×1$
Apply distributive property for first two terms and last two terms,
$5x×x+1×5x-3×x-3×1=5x\left(x+1\right)-3\left(x+1\right)$
Again apply distributive property for $5x\left(x+1\right)-3\left(x+1\right)$
$5x\left(x+1\right)-3\left(x+1\right)=\left(5x-3\right)\left(x+1\right)$
So the factors are $\left(5x-3\right)$ and $\left(x+1\right)$
$5{x}^{2}+2x-3=\left(5x-3\right)\left(x+1\right)$
To find the value of x,
Take $x+1=0$
Subtract 1 on both sides,
$x+1-1=0-1$
$x=-1$
Take $5x-3=0$
$5x-3+3=0+3$
$5x=3$
Divide by 5 on both sides,
$\frac{5x}{5}=\frac{3}{5}$
$x=\frac{3}{5}$
So, the values of x are $\frac{3}{5}$ and -1

Annie Gonzalez

Lets

karton

Step 1
Given: x(5x+2)=3
Expand:
Apply the distributive law: a(b+c)=ab+ac

a=x, b=5x, c=2

x*5x+x*2
=5x* x+2x

Apply exponent rule: ${a}^{b}×{a}^{c}={a}^{b+c}$$x×x={x}^{1+1}$
$=5{x}^{1+1}$
$=5{x}^{2}$
$=5{x}^{2}+2x$
$5{x}^{2}+2x=3$
Subtract 3 from both sides
$5{x}^{2}+2x-3=3-3$
Simplify
$5{x}^{2}+2x-3=0$

For a=5, b=2, c=-3

$\sqrt{{2}^{2}-4×5\left(-3\right)}$
Apply rule -(-a)=a
$=\sqrt{{2}^{2}+4×5×3}$
Multiply the numbers: $4×5×3=60$
$=\sqrt{{2}^{2}+60}$
${2}^{2}=4$
$=\sqrt{4+60}$
Add the numbers: $4+60=64$
$=\sqrt{64}$
Factor the number: $64={8}^{2}$
$=\sqrt{{8}^{2}}$
Apply radical rule: $\sqrt[n]{{a}^{n}}=a$
$\sqrt{{8}^{2}}=8$
$=8$

Separate the solutions

$x=\frac{-2+8}{2×5}:\phantom{\rule{1em}{0ex}}\frac{3}{5}$

The solutions to the quadratic equation are:

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