Question

Determine the number and type of solutions for x2 + 9x + 7 = 0.

Answer 1

that equation is in the form:\[ax ^{2} + bx + c = 0\]

So, you have a=1, b=9, and c=7 Your solutions will be irrational and will have the form:\[\frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a }\]

Answer 2

So, from the above expression I gave you, there is a "+-" sign. which means one "x" will correspond to the "+" and the other "x" will correspond to the "-" sign. So, just make the substitutions and simplify.

Answer 3

Can you get the b^2 - 4ac ?

Answer 4

Your highest power of x is x^2, so this is a second degree polynomial. The fundamental theorem of algebra says that you'll have 2 roots (aka zeros, aka solutions) for a polynomial of degree 2.

With Descartes' rule of signs, we can determine the nature of the roots. With the equation written in descending exponent order (as you have it), count the number of changes of sign. Here each term is positive, so there are no changes of sign. There will be at most (count of sign changes) positive roots, although the number may be less by a multiple of 2.

Next, change the sign of any terms with odd exponents (x, x^3, x^5, etc.) and redo the count of sign changes. Our new polynomial would be x^2-9x+7=0 so the count would be 3. There are at most (count of sign changes) negative roots, again possibly reduced by a multiple of 2.

Finally, complex roots make up the remainder of the roots, and always come in pairs if the coefficients of the polynomial are real numbers.

So, we had no sign changes for the original polynomial -> 0 positive roots
We had 3 sign changes for the altered polynomial -> 2 negative roots, because there are only 2 roots in a 2nd degree polynomial
No leftover roots to be complex roots.

Answer 5

b^2 means "b squared"

Answer 6

Sorry, the count in the altered equation is 2, not 3. There were 3 different signs, but only 2 sign changes.

Answer 7

The b^2 - 4ac expression relates to the discriminant. You will see that this is a positive number, so your 2 roots will be real, but not rational. You will see this because the square root of this expression cannot be put into form p/q where p and q are integers. You will have 2 real roots, both irrational.

Answer 8

This parabola opens upward and the 2 real, irrational zeros imply that the graph crosses the x-axis twice. You do not need a graph for your answer, but the graph illustrates that real solutions cross the x-axis.

Answer 9

If the discriminant is > 0, you get 2 real roots. If the discriminant is < 0, you get 2 complex roots. If the discriminant = 0, then you get a double root. An example of the latter would be

\[(x-2)(x-2) = 0\]
\[(x-2)(x-2) = x^2-4x+4\]
\[a=1, b=-4, c = 4\]
\[x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2(1)} = \frac{4\pm\sqrt{16-16}}{2} = 2\]

Answer 10

btw, those 2 real irrational roots are both negative, just in case you need to know the sign of the roots. You can tell that by the graph where both crossings of the x-axis are to the left of x = 0.