rehaucesnwr1

2023-02-22

How to solve $5{x}^{2}-3x=25$ by completing the square?

Jayden Landry

So solution:
$5{x}^{2}-3x=25\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}5\left({x}^{2}-\frac{3}{5}x\right)=25$
$5\left({x}^{2}-\frac{3}{5}x+\frac{9}{100}\right)=25+\frac{9}{20}$
$5{\left(x-\frac{3}{10}\right)}^{2}=\frac{509}{20}$
${\left(x-\frac{3}{10}\right)}^{2}=\frac{509}{100}$
${\left(x-\frac{3}{10}\right)}^{2}=\frac{509}{100}$
$x=0.3±2.256\therefore x=2.556,x=-1.956$
Result: $x\approx 2.556,x\approx -1.956$

Libby Fernandez

Here is an example of how to solve using the difference of squares identity, premultiplying by 20 to reduce the number of fractions, and completing the square:
${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$
with #a=(10x-3)# and #b=sqrt(509)#
We have:
$5{x}^{2}-3x=25$
25 from each side is subtracted to yield:
$5{x}^{2}-3x-25=0$
Multiply by $20=5\cdot {2}^{2}$ to make the leading term a perfect square and the middle term divisible by 2...
Therefore:
$0=20\left(5{x}^{2}-3x-25\right)$
$0=100{x}^{2}-60x-500$
$0={\left(10x\right)}^{2}-2\left(10x\right)\left(3\right)+{3}^{2}-509$
$0={\left(10x-3\right)}^{2}-{\left(\sqrt{509}\right)}^{2}$
$0=\left(\left(10x-3\right)-\sqrt{509}\right)\left(\left(10x-3\right)+\sqrt{509}\right)$
$0=\left(10x-3-\sqrt{509}\right)\left(10x-3+\sqrt{509}\right)$
Hence:
$10x=3±\sqrt{509}$
So:
$x=\frac{3}{10}±\frac{\sqrt{509}}{10}$

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