Question

Solve using quadratic formula for 2x^(2)-10x+5=0

Answer 1

`x=(5+√(15))/(2),\:x=(5-√(15))/(2)`

Explanation:

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}`

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=2,\:b=-10,\:c=5`

`x_(1,\:2)=(-\left(-10\right)\pm √(\left(-10\right)^2-4\cdot \:2\cdot \:5))/(2\cdot \:2)`

`√(\left(-10\right)^2-4\cdot \:2\cdot \:5)`

Apply exponent rule: (-a)^n=a^n, if n is even

`\left(-10\right)^2=10^2`

`=√(10^2-4\cdot \:2\cdot \:5)`

`\mathrm{Multiply\:the\:numbers:}\:4\cdot \:2\cdot \:5=40`

`=√(10^2-40)`

`10^2=100`

`=√(100-40)`

`\mathrm{Subtract\:the\:numbers:}\:100-40=60`

`=√(60)`

`60\:\mathrm{divides\:by}\:2\quad \:60=30\cdot \:2`

`=2\cdot \:30`

`30\:\mathrm{divides\:by}\:2\quad \:30=15\cdot \:2`

`=2\cdot \:2\cdot \:15`

`15\:\mathrm{divides\:by}\:3\quad \:15=5\cdot \:3`

`=2\cdot \:2\cdot \:3\cdot \:5`

`2,\:3,\:5\mathrm{\:are\:all\:prime\:numbers,\:therefore\:no\:further\:factorization\:is\:possible}`

`=2\cdot \:2\cdot \:3\cdot \:5`

`=2^2\cdot \:3\cdot \:5`

`=√(2^2\cdot \:3\cdot \:5)`

Apply Radical Rule:

`=√(2^2)√(3\cdot \:5)`

Apply Radical Rule:

`√(2^2)=2`

`=2√(3\cdot \:5)`

`\mathrm{Refine}`

`=2√(15)`

`x_(1,\:2)=(-\left(-10\right)\pm \:2√(15))/(2\cdot \:2)`

`\mathrm{Separate\:the\:solutions}`

`x_1=(-\left(-10\right)+2√(15))/(2\cdot \:2),\:x_2=(-\left(-10\right)-2√(15))/(2\cdot \:2)`

`(-\left(-10\right)+2√(15))/(2\cdot \:2)`

Apply Rule -(-a)=a

`=(10+2√(15))/(2\cdot \:2)`

`\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4`

`=(10+2√(15))/(4)`

`=(2\left(5+√(15)\right))/(4)`

`\mathrm{Cancel\:the\:common\:factor:}\:2`

`=(5+√(15))/(2)`

`(-\left(-10\right)-2√(15))/(2\cdot \:2)`

`=(10-2√(15))/(2\cdot \:2)`

`\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4`

`=(10-2√(15))/(4)`

`=(2\left(5-√(15)\right))/(4)`

`\mathrm{Cancel\:the\:common\:factor:}\:2`

`=(5-√(15))/(2)`

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

`x=(5+√(15))/(2),\:x=(5-√(15))/(2)`