 # What are the roots of the equation x^2-5x+1=0 ? guringpw 2021-12-15 Answered
What are the roots of the equation ${x}^{2}-5x+1=0$ ?
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You can solve this equation using 2 methods, one being completing the square method, and the other by using the quadratic formula.
1) Commencing completing the square method,
${x}^{2}-5x+1=0$
Subtract 1 on both sides,
${x}^{2}-5x=-1$
Add 6.25 on both sides,
${x}^{2}-5x+6.25=-1+6.25$
Apply perfect quadratic square formula,
${\left(x-2.5\right)}^{2}=5.25$
Square root both sides,
$x-2.5=±\sqrt{5.25}$
Add 2.5 to both sides,
$x=±\sqrt{5.25}+2.5$
Hence,
$x=4.79128784$ or $0.20871215$
2) Commencing quadratic formula method,
${x}^{2}-5x+1=0$
$a{x}^{2}+bx+c=0$
Substitute $a=1,b=-5,c=1$ into the quadratic formula.
$x=\frac{5±\sqrt{21}}{2}$
Hence,
$x=4.79128784$ or $0.20871215$
###### Not exactly what you’re looking for? censoratojk
Find the roots:
${x}^{2}-5x+1=0⇒$ quadratic equation
The standard form for a quadratic equation is $a{x}^{2}+bx+c=0$, where $a=1,b=-5$ and $c=1$. Note: $a\ne 0$
Solve this quadratic equation using the quadratic formula:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Substitute the known values into the formula.
$x=\frac{-\left(-5\right)±\sqrt{{\left(-5\right)}^{2}-4\cdot 1\cdot 1}}{2\cdot 1}$
Simplify.
$x=\frac{5±\sqrt{25-4}}{2}$
Simplify.
$x=\frac{5±\sqrt{21}}{2}$
Solutions for x. roots
$x=\frac{5+\sqrt{21}}{2},\frac{5-\sqrt{21}}{2}$