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Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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Section
BuyFindarrow_forward

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

With the highest degree term being x 4 , the equation x 4 - 13 x 2 + 36 = 0 is known as a quartic equation. Use factoring to find four real solutions for this equation.

To determine

To solve:

The quadratic equation by factorization.

x4=13x2+36=0

Explanation

Approach:

Solving an equation is to find the value of the unknown variables in the equation, such that the obtained value or values of the unknown should satisfy the equation from which it was derived. Such a value is said to be the solution for the equation. In general a quadratic equation has two solutions for the variable in the equation as the degree of the equation is two.

Calculation:

Given,

x4=13x2+36=0

x22-13x2+36=0

Let us assume that x2=u.

u2-13u+36=0

The above equation is a quadratic equation.

The above equation is in the standard form of a quadratic equation.

Now, we shall proceed with the factorization of the quadratic expression in the left of the equation.

First look for the common factor GCF of the terms u2, -13u and 36.

But, here the GCF is 1.

Thus, we have to factor the above equation by the reverse FOIL method.

First factorize the product of the coefficient of the first tern and the constant, such that the sum of the factors gives the coefficient of the middle term.

Thus, factorize 36 as -9·-4 so that the sum of the factors -9+(-4) gives -13.

Now, split the middle term -13u as -9u+(-4u).

Hence, the equation becomes,

u2+-9u+-4u+36=0

u·u+u·-9+u·-4+-9·-4=0

Factor out the common factors form the first two terms and form the last two terms of the above expression

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