Question

Find the zeros of quadratic polynomial 7x2 – 6x – 16.

Answer 1

Answer:

7x2+6x-16=0 

Two solutions were found :

 x = -2

 x = 8/7 = 1.143

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(7x2 + 6x) - 16 = 0

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  7x2+6x-16 

The first term is,  7x2  its coefficient is  7 .

The middle term is,  +6x  its coefficient is  6 .

The last term, "the constant", is  -16 

Step-1 : Multiply the coefficient of the first term by the constant   7 • -16 = -112 

Step-2 : Find two factors of  -112  whose sum equals the coefficient of the middle term, which is   6 .

     -112   +   1   =   -111     -56   +   2   =   -54     -28   +   4   =   -24     -16   +   7   =   -9     -14   +   8   =   -6     -8   +   14   =   6   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -8  and  14 

                     7x2 - 8x + 14x - 16

Step-4 : Add up the first 2 terms, pulling out like factors :

                    x • (7x-8)

              Add up the last 2 terms, pulling out common factors :

                    2 • (7x-8)

Step-5 : Add up the four terms of step 4 :

                    (x+2)  •  (7x-8)

             Which is the desired factorization

Equation at the end of step  2  :

(7x - 8) • (x + 2) = 0

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2      Solve  :    7x-8 = 0 

 Add  8  to both sides of the equation : 

                      7x = 8

Divide both sides of the equation by 7:

                     x = 8/7 = 1.143

Solving a Single Variable Equation :

 3.3      Solve  :    x+2 = 0 

 Subtract  2  from both sides of the equation : 

                      x = -2

Answer 2

Answer:

$7 {x}^{2} - 6x - 16 = 0 \\ let \: zeros \: of \: quadratic \: equation \: are \: \alpha \: \: and \: \beta \\ = > \alpha + \beta = \frac{ - b}{a} \\ \alpha \beta = \frac{c}{a} \\ \\ here \: a = 7 \\ b = - 6 \\ c = - 16 \\ \\ \alpha + \beta = - \frac{( - 6)}{7} \\ = > \alpha + \beta = \frac{6}{7} \\ \\ \alpha \beta = \frac{ - 16}{7} \\ = > \beta = \frac{ - 16}{7 \ \alpha } \\ \\ = > \alpha - \frac{16}{7 \ \alpha } = \frac{6}{7} \\ = > \frac{7 { \alpha }^{2} - 16 }{7 \alpha } = \frac{6}{7 } \\ = > 7 { \alpha }^{2} - 16 = \frac{42 \alpha }{7} \\ = > 7 { \alpha }^{2} - 16 = 6 \alpha \\ = > 7 { \alpha }^{2} - 6 \alpha - 16 = 0 \\ = > \alpha = \frac{6 + - \sqrt{ {( - 6)}^{2} } + 4 \times 7 \times 16}{2 \times 7} \\ = > \alpha = \frac{6 + - \sqrt{36} + 448 }{14} \\ = > \alpha = \frac{6 + - \sqrt{484} }{14} \\ = > \alpha = \frac{6 + - 22}{14} \\ = > \alpha = \frac{6 + 22}{14} \: \frac{6 - 22}{14} \\ = > \alpha = \frac{28}{14} \: \: . \frac{ - 16}{14} \\ \alpha = 2 \: \: and \: \: \: \frac{ - 8}{7 } \\ \\ \\ \alpha \beta = \frac{ - 16}{7} \\ = > 2 \times \beta = \frac{ - 16}{7 } \\ = > \beta = \frac{ - 16}{14}$