Question

Find the zeros of the function: to the nearest hundredth

Answer 1

The zeros of the function:

`x=5.53 ,`
`x=3.22`

Explanation:

Given the function

`f\left(x\right)=2x^2-17.5x+35.6`

To find zeros, put f(x)=0

`0=2x^2-17.5x+35.6`

`\mathrm{Multiply\:both\:sides\:by\:}10`

`0\cdot \:10=2x^2\cdot \:10-17.5x\cdot \:10+35.6\cdot \:10`

`0=20x^2-175x+356`

switch sides

`20x^2-175x+356=0`

subtract 356 from both sides

`20x^2-175x+356-356=0-356`

`20x^2-175x=-356`

divide both sides by 20

`(20x^2-175x)/(20)=(-356)/(20)`

`x^2-(35x)/(4)=-(89)/(5)`

`\mathrm{Add\:}a^2=\left(-(35)/(8)\right)^2\mathrm{\:to\:both\:sides}`

`x^2-(35x)/(4)+\left(-(35)/(8)\right)^2=-(89)/(5)+\left(-(35)/(8)\right)^2`

applying perfect square

`\left(x-(35)/(8)\right)^2=(429)/(320)`

`\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=√(a),\:-√(a)`

solving

`x-(35)/(8)=\sqrt{(429)/(320)}`

`x=(√(2145))/(40)+(35)/(8)`

`x=5.53`

also solving

`x-(35)/(8)=-\sqrt{(429)/(320)}`

`x=-(√(2145))/(40)+(35)/(8)`

`x=3.22`

`\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}`

`x=5.53 ,`
`x=3.22`

Therefore, the zeros of the function:

`x=5.53 ,`
`x=3.22`