Question

Question #e94e0

Answer 1

`r=7 or r =2`

Explanation:

`color(blue)(r^2=-14+9r`

Bring everything to the left hand side

`rarrr^2-9r+14=0`

Now, this is a quadratic equation. Solve it by using the Quadratic formula

`color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)`

Where `x` is the variable `r` and `a,band c` are the coefficients of the terms `(a=1,b=-9 and c=14)`

`rarrr=(-(-9)+-sqrt((-9)^2-4(1)(14)))/(2(1))`

`rarrr=(9+-sqrt(81-56))/(2)`

`rarrr=(9+-sqrt(25))/(2)`

`rarrr=(9+-5)/(2)`

Now, there are two solutions for `r`

`color(purple)(r=(9+5)/2=14/2=7)`

`color(violet)(r=(9-5)/2=4/2=2)`

`color(green)( :.r=7` `color(green)(or` `color(green)(2`

Hope this helps..! :)

Answer 2

`r = 2 or r =7`

Explanation:

Re-arrange it into general form first: `ax^2 +bx +c=0`

It does not matter at all what the variable is!

`r^2 -9r +14 =0`

This means: `a = 1" "b = -9" " c =14`

The quadratic formula is: `x = (-b +- sqrt(b^2 -4ac))/(2a)`

Use the values for `a,b,c` above and substitute.

`r = (-(-9) +- sqrt((-9)^2 -4(1)(14)))/(2(1))" "larr` simplify

`r = (9 +- sqrt((81 -56)))/2`

There are two possible answers for `r`

`r = (9+sqrt25)/2 = 7`

`r = (9-sqrt25)/2 = 2`

We could also have solved the original equation by finding the factors.

Answer 3

`r=2" or " r=7`

Explanation:

`"rearrange and equate to zero"`

`rArrr^2-9r+14=0`

`"compare to the standard form " ax^2+bx+c=0`

`"here " a=1, b=-9, c=14`

`rArrr=(-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(rArrr)=(-(-9)+-sqrt((-9)^2-(4xx1xx14)))/2`

`color(white)(rArrr)=(9+-sqrt(81-56))/2=(9+-sqrt25)/2`

`rArrr=(9+5)/2=7" or " r=(9-5)/2=2`