+0

+1
250
3
+1405

Exercise:

Are the roots real or imaginary?

5x2-7x-6=0

Mar 6, 2018

#2
+2339
+1

The question asks about the type of roots--not what the roots are.

Because of this technicality, we can combine knowledge from Descartes' Rule of Signs and the Fundamental Theorem of Algebra and the Complex Conjugate Root Theorem in order to answer this question.

The Descartes' Rule of Signs states that the number of sign changes in $$f(x)$$, when the terms are ordered by degree, indicates the number of positive solutions or less than this by an even number. Let's consider the original function $$f(x)=5x^2-7x-6$$:

$$f(x)=\underbrace{5x^2-7}\underbrace{x-6}\\ \hspace{16mm}+1\hspace{7mm}+0\hspace{10mm}=\text{1 sign change}$$

Notice that +5x2 and -7x have a different sign, so I marked that change with a +1. -7x and -6 do not have a signage change, so I marked it with a +0.

The Fundamental Theorem of Algebra is relevant because it indicates how many total solutions that a function can have. In this case, there are 2 roots since the degree of the polynomial is 2.

The Complex Conjugate Root Theorem states that if polynomials have real coefficients with complex roots, then they will come in pairs; in other words, there will always be an even number of non-real roots.

I would say that it would be wise to consolidate this information into a table.

 Number of Positive Solutions Number of Negative Solutions Number of Complex Roots Total Number of Roots 1 1 0 2

Based on the rules mentioned previously, only one possibility remains: Both roots are real.

Mar 7, 2018

#1
+68
+1

$$5x^2-7x-6=0$$

First, do grouping.

$$5x^2-10x+3x-6=0$$

$$5x(x-2)+3(x-2)=0$$

$$(5x+3)(x-2)=0$$

Then "tee-up" and solve for $$x$$.

$$5x+3=0, x-2=0$$

$$x=-\frac{3}{5}, x=2$$

The roots are very much so real.

Mar 6, 2018
#2
+2339
+1

The question asks about the type of roots--not what the roots are.

Because of this technicality, we can combine knowledge from Descartes' Rule of Signs and the Fundamental Theorem of Algebra and the Complex Conjugate Root Theorem in order to answer this question.

The Descartes' Rule of Signs states that the number of sign changes in $$f(x)$$, when the terms are ordered by degree, indicates the number of positive solutions or less than this by an even number. Let's consider the original function $$f(x)=5x^2-7x-6$$:

$$f(x)=\underbrace{5x^2-7}\underbrace{x-6}\\ \hspace{16mm}+1\hspace{7mm}+0\hspace{10mm}=\text{1 sign change}$$

Notice that +5x2 and -7x have a different sign, so I marked that change with a +1. -7x and -6 do not have a signage change, so I marked it with a +0.

The Fundamental Theorem of Algebra is relevant because it indicates how many total solutions that a function can have. In this case, there are 2 roots since the degree of the polynomial is 2.

The Complex Conjugate Root Theorem states that if polynomials have real coefficients with complex roots, then they will come in pairs; in other words, there will always be an even number of non-real roots.

I would say that it would be wise to consolidate this information into a table.

 Number of Positive Solutions Number of Negative Solutions Number of Complex Roots Total Number of Roots 1 1 0 2

Based on the rules mentioned previously, only one possibility remains: Both roots are real.

TheXSquaredFactor Mar 7, 2018
#3
+1405
+1

Thanks Guys

Mar 7, 2018