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2023-02-22

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

Jayden Landry

Beginner2023-02-23Added 7 answers

So solution:

$5{x}^{2}-3x=25\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}5({x}^{2}-\frac{3}{5}x)=25$

$5({x}^{2}-\frac{3}{5}x+\frac{9}{100})=25+\frac{9}{20}$

$5{(x-\frac{3}{10})}^{2}=\frac{509}{20}$

$(x-\frac{3}{10})}^{2}=\frac{509}{100$

$(x-\frac{3}{10})}^{2}=\frac{509}{100$

$x=0.3\pm 2.256\therefore x=2.556,x=-1.956$

Result: $x\approx 2.556,x\approx -1.956$

$5{x}^{2}-3x=25\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}5({x}^{2}-\frac{3}{5}x)=25$

$5({x}^{2}-\frac{3}{5}x+\frac{9}{100})=25+\frac{9}{20}$

$5{(x-\frac{3}{10})}^{2}=\frac{509}{20}$

$(x-\frac{3}{10})}^{2}=\frac{509}{100$

$(x-\frac{3}{10})}^{2}=\frac{509}{100$

$x=0.3\pm 2.256\therefore x=2.556,x=-1.956$

Result: $x\approx 2.556,x\approx -1.956$

Libby Fernandez

Beginner2023-02-24Added 5 answers

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:

${a}^{2}-{b}^{2}=(a-b)(a+b)$

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:

$0=20(5{x}^{2}-3x-25)$

${0}=100{x}^{2}-60x-500$

${0}={\left(10x\right)}^{2}-2\left(10x\right)\left(3\right)+{3}^{2}-509$

$0}={(10x-3)}^{2}-{\left(\sqrt{509}\right)}^{2$

${0}=((10x-3)-\sqrt{509})((10x-3)+\sqrt{509})$

${0}=(10x-3-\sqrt{509})(10x-3+\sqrt{509})$

Hence:

$10x=3\pm \sqrt{509}$

So:

$x=\frac{3}{10}\pm \frac{\sqrt{509}}{10}$

${a}^{2}-{b}^{2}=(a-b)(a+b)$

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:

$0=20(5{x}^{2}-3x-25)$

${0}=100{x}^{2}-60x-500$

${0}={\left(10x\right)}^{2}-2\left(10x\right)\left(3\right)+{3}^{2}-509$

$0}={(10x-3)}^{2}-{\left(\sqrt{509}\right)}^{2$

${0}=((10x-3)-\sqrt{509})((10x-3)+\sqrt{509})$

${0}=(10x-3-\sqrt{509})(10x-3+\sqrt{509})$

Hence:

$10x=3\pm \sqrt{509}$

So:

$x=\frac{3}{10}\pm \frac{\sqrt{509}}{10}$