+0

Algebra

0
135
2

Find all solutions to the equation x^2 + 29 = 10x + 3.

Jun 25, 2021

#1
+152
+2

$x^2+29=10x+3$

take the 3 to the left-hand side

$x^2+26=10x$

now take the 10x to the other side

$x^2-10x+26=0$

now to solve this you can pick many ways, though i will pick using the quadratic formula

$_2x_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where $a=1$  ;  $b=-10$ ;  $c=26$

just plug and chug

$_2x_1=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:26}}{2\cdot \:1}$

$_2x_1=\frac{-\left(-10\right)\pm \iota \sqrt{10^2-104}} {2\cdot \:1}$

$_2x_1=\frac{-\left(-10\right)\pm \iota \sqrt{4} } {2\cdot \:1}$

$_2x_1=\frac{-\left(-10\right)\pm \:2\iota }{2\cdot \:1}$

that gives us $x_1=\frac{10+2\iota }{2\cdot \:1} \ \ \ \Rightarrow \ \ \ \frac{10+2\iota }{2} \ \ \ \Rightarrow \ \ \ \frac{\cancel{2}\left(5+\iota \right)} {\cancel{2}}$

finally, we get $x_1==5+\iota$

second root will be as following:

$x_2=\frac{10-2\iota }{2\cdot \:1} \ \ \ \Rightarrow \ \ \ \frac{10-2\iota a}{2} \ \ \ \Rightarrow \ \ \ \frac{\cancel{2}\left(5-\iota \right)}{\cancel{2}}$

thus the second root is $x_2=5-\iota$

Jun 25, 2021
#2
+152
+2

that gives us $x_1=\frac{10+2\iota }{2\cdot \:1} \ \ \ \Rightarrow \ \ \ \frac{10+2\iota }{2} \ \ \ \Rightarrow \ \ \ \frac{\cancel{2}\left(5+\iota \right)}{ \cancel{2}}$

finally, we get

$x_1=5+\iota$

fixed.