x^2-15x+54=155 solving by rational root theorem, descartes rule of signs, and the factor theorem.

Question

solving by rational root theorem, descartes rule of signs, and the factor theorem.

Answer 1

Given polynomial is x^2-15x+54 = 155

Subtract 155 from each side.

x^2-15x+54-155 = 0

x^2-15x-101 = 0

By rational root therom

x^2-15x-101 = 0

Since leading coefficient is 1, any rational zero must be a divisor of the constant term101.

So the possible rational zeros are +1,-1,+101,-101.

We test each of possibilties.

P(1) = 1-15-101 = -115 not equls to 0.

P(-1) = 1+15-101 = 85 not equals to 0.

P(101) = 101^2-15*101-101 = 10201-1515-101 = 8585 not equals to 0.

p(-101) = (-101)^2-15*-101-101 = 10201+1515-101 = 13130 not equals to 0.

By descartes rule of signs

f(x) = x^2-15x-101

In the positive case sign changes one time there is one positive root.

f(-x) = x^2+15x-101

In the negitive case sign changes one time there is one negitive root.

x^2-15x-101 = 0

Now factorize x^2-15x-101 = 0

Compare it to standard quadratic equation ax^2+bx+c = 0

a = 1, b = -15, c = -101

we know that x = [-b+sqrt(b^2-4ac)]/2a   and      x = [-b-sqrt(b^2-4ac)]/2a

x = [-(-15)+sqrt(225-4*1*-101)]/2*1       and      x = [-(-15)-sqrt(225-4*1*-101)]/2*1

x = 15+sqrt(225+404)/2                           and     x = 15-sqrt(225+404)/2

x = (15+sqrt(629))/2                                 and     x = (15-sqrt(629))/2

x = (15+25.07)/2                                      and     x = (15-25.07)/2

x = 20.03                                                 and      x = -5.03

Solution x = 20.03 and x = -5.03