   Chapter 6.4, Problem 59PS ### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

#### Solutions

Chapter
Section ### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

# For each quadratic equation in Problems 51 − 60 , first use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation. (Objective 2) 3 x 2 + 4 x = 2

To determine

To solve:

The given quadratic equation.

Explanation

The general form of quadratic equation is ax2+bx+c=0.

The property of factorization is ab=0 if and only if a=0 and b=0.

If D<0, then the equation has two non real complex solutions.

If D>0, then the equation has two real solutions.

If D=0, then the equation has one real solution with multiplicity two.

D is the discriminant of the equation.

Given:

The given quadratic equation is,

3x2+4x=2

Formula used:

The solution of the equation ax2+bx+c=0 is given by,

x=b±D2a

Where D is the discriminant and is given by,

D=b24ac

Where, a and b are the coefficients of x2andx respectively. c is the constant term for the quadratic equation ax2+bx+c=0.

Calculation:

The given quadratic equation is,

3x2+4x=2(1)

Equation (1) can be rewritten as,

3x2+4x2=0.... (2)

Compare the equation (2) with the general form of quadratic equation, the values of a, b, and c are 3, 4, and 2 respectively.

Substitute 3 for a, 4 for b, and 2 for c in the above mentioned formula for discriminant.

D=(4)24(3)(2)=16+24=40

The equation has two real solutions as the value of discriminant is positive

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