Question

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Answer 1

D

hope this helps :o)

Answer 2

Hey there, hope I can help!


`\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}`

`2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)`

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request

`x^2-2x-46=0`

Lets use the quadratic formula now

`\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\:} a=1,\:b=-2,\:c=-46: x_(1,\:2)=(-\left(-2\right)\pm √(\left(-2\right)^2-4\cdot \:1\left(-46\right)))/(2\cdot \:1)`


`(-\left(-2\right)+√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)))/(2\cdot \:1) \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ (2+√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)))/(2\cdot \:1)`

Multiply the numbers 2 * 1 = 2

`(2+√(\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4))/(2)`


`2+√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)) \ \textgreater \ √(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right))`


`\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ √(\left(-2\right)^2+1\cdot \:4\cdot \:46) \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4`


`\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ √(4+184) \ \textgreater \ √(188) \ \textgreater \ 2 + √(188)`

`(2+√(188))/(2) \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ √(2^2\cdot \:47)`


`\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ √(47)√(2^2)`


`\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ √(2^2)=2 \ \textgreater \ 2√(47) \ \textgreater \ (2+2√(47))/(2)`


`Factor\;2+2√(47) \ \textgreater \ Rewrite\;as\;1\cdot \:2+2√(47)`

`\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+√(47)\right) \ \textgreater \ (2\left(1+√(47)\right))/(2)`


`\mathrm{Divide\:the\:numbers:}\:(2)/(2)=1 \ \textgreater \ 1+√(47)`

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.


`(-\left(-2\right)-√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)))/(2\cdot \:1) \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ (2-√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)))/(2\cdot \:1)`


`(2-√(\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4))/(2)`


`2-√(\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)) \ \textgreater \ 2-√(188) \ \textgreater \ (2-√(188))/(2)`


`√(188) = 2√(47) \ \textgreater \ (2-2√(47))/(2)`


`2-2√(47) \ \textgreater \ 2\left(1-√(47)\right) \ \textgreater \ (2\left(1-√(47)\right))/(2) \ \textgreater \ 1-√(47)`

Therefore our final solutions are

`x=1+√(47),\:x=1-√(47)`

Hope this helps!