Solve the equation by completing the square. 2x-18x+58=0

1 answer:

5 0

$x^{2} -18x+58=0$

We have got quadratic equation. We calculate it by using Δ.

$\Delta=(-18)^{2} -4*1*58=324-232=92>0$
$\sqrt{\Delta} = \sqrt{92} =2 \sqrt{23}$

$x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{18-2 \sqrt{23} }{2} =9- \sqrt{23}$
∨
$x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{18+2 \sqrt{23} }{2} =9+ \sqrt{23}$

Hope it helps :)

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Step-by-step explanation:

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Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

Equation Solving

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$