This Math Question Looks Easy… Until You Try It! #maths #math

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#brainexercise #matholympics
This Math Question Looks Easy… Until You Try It! #maths #math

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This can be solved much easier. We are given that

x² + 3x − 7 = 0

and need to evaluate

x + 3/(x + 4)

The expression to evaluate looks like the result of a polynomial division of a quadratic polynomial by x + 4 with a remainder. Now, dividing x² + 3x − 7 by x + 4 we have

(x² + 3x − 7)/(x + 4) = x − 1 − 3/(x + 4)

which means that for the values of x that satisfy the given quadratic equation we have

0 = x − 1 − 3/(x + 4)

and therefore

3/(x + 4) = x − 1

so

x + 3/(x + 4) = 2x − 1

and since the roots of the quadratic equation are x = −³⁄₂ ± ¹⁄₂√37 it immediately follows that

x + 3/(x + 4) = −4 ± √37

Most of these types of problems are created in such a way that the expression to evaluate has a single value for each of the values that satisfy the given equation so, as other commentators have also observed, one may wonder whether perhaps the original problem asked to evaluate

x − 3/(x + 4)

which of course equals 1 for each of the two roots of the given quadratic equation.

NadiehFan
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(x+4)(x-1) = x^2+3x-4 = 3 after substituting with the first equation.
So x-1 = 3 / (x+4).
Result is x + x-1 = 2x -1
Since x = (-3 +- sqrt(37)) / 2, final result is -4 +- sqrt(37)

sasak
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Are you sure that you want ask what you ask? :)

If you would ask the value of x-3/(x+4) than you will get 1 that much more prettier than irrational radical mess.

ald
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D= (-3)^2-4*1*(-7) = 9+28 = 37 so x= (-3+ sqrt 37) / 2. Then x + (3/x+4) = x * (x+4) / (x+4)+3 / (x+4) =
( x^2 + 4x +3 ) / ( x + 4 ) =
(x^2+3x+x - 7+10) / (x+4) = (0+x+10) / (x+4) = (x+10) / (x+4) = (x+4+6) / (x+4) = 1+6 / (x+4) =
1+6 / ((-3+ sqrt37)/2) + (8/2 )= 1+12 / (5 + - sqrt37 ) = 1+12 * ( 5 - +sqrt37) / ((5+ - sqrt37) * (5 - +sqrt37) = 1+12* (5 - +sqrt37) / (25-37) = 1+ 12* ( 5 - +sqrt37) / (-12) = 1 - (5- +sqrt37) = 1-5+ -sqrt37= - 4+ - sqrt37

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