Question #e94e0

3 Answers
Apr 27, 2017

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..! :)

Apr 27, 2017

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.

Apr 27, 2017

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