Question

Solve the equation x2-16x+54=0

Answer 1

Answer:

x1 = 8+√10, x2 = 8 -√10

Step-by-step explanation:

x² - 16x + 54 = 0

$x = \frac{-b +/-\sqrt{b^{2} -4ac} }{2a} \\x = \frac{-(-16)+/-\sqrt{(-16)^{2} -4*1*54} }{2*1} \\\\\\x = \frac{16+/-\sqrt{(40)} }{2*1} = 8+/-\sqrt{10} \\\\x_{1} = 8+\sqrt{10}\\\\x_{2} =8-\sqrt{10}$

Answer 2

The value of the x is x=8+root 10,and \:x=8-root 10.

What is the formula of the quadratic equation?

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore by using the formula we have

$x^2- 16x + 54 = 0$

a=1,b=-16 and c=54

Therefore we get,

$x_{1,\:2}=\frac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\cdot \:1\cdot \:54}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-\left(-16\right)\pm \:2\sqrt{10}}{2\cdot \:1}$

$x_1=\frac{-\left(-16\right)+2\sqrt{10}}{2\cdot \:1},\:x_2=\frac{-\left(-16\right)-2\sqrt{10}}{2\cdot \:1}$

$x=8+\sqrt{10},\:x=8-\sqrt{10}$

Therefore we get the roots of the given polynomials are

$x=8+\sqrt{10},and \:x=8-\sqrt{10}$

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