How do you solve $v^{2} - 95 v - 96 = 0$ $v^ { 2} - 95v - 96= 0$ ?

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How do you solve $v^{2} - 95 v - 96 = 0$ $v^ { 2} - 95v - 96= 0$ ?

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Answer 1 · 2017-12-07T07:10:16

$v = 96, v = - 1$ $color(blue)(v = 96, v = -1)$

Explanation:

We are given the quadratic equation:

$v^{2} - 95 v - 96 = 0$ $color(blue)(v^2-95v-96) = 0$ $. . E q u a t i o n .1$ $.. color(red)(Equation.1)$

Our quadratic expression is

$v^{2} - 95 v - 96$ $color(blue)(v^2-95v-96)$ $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$

To factor this quadratic expression, we will follow the procedure given below:

$S t e p .1$ $color(green)(Step.1)$

We must split the coefficient of middle term into two numbers , such that when we add them we get the middle term, and when we multiply them we must get the product of the coefficient of the $x^{2} t e r m$ $x^2 term$ and the constant,

Note that the product of the coefficient of the $x^{2} t e r m$ $x^2 term$ and the constant is $(- 96)$ $(-96)$ ,

$S t e p .2$ $color(green)(Step.2)$

The two numbers are: $- 96 and + 1$ $color(blue)(-96 and +1)$

When we add ( - 96) and ( +1 ) we get $(- 95)$ $(- 95)$ and when we multiply the two values ( - 96) and ( +1 ) we get ( - 96 )

Now, we write our $. . E x p r e s s i o n .1$ $.. color(red)(Expression.1)$ as follows:

$v^{2} - 96 v + 1 v - 96$ $color(blue)(v^2-96v + 1v-96)$ $. . E x p r e s s i o n .2$ $.. color(red)(Expression.2)$

$S t e p .3$ $color(green)(Step.3)$

In this step, we break our $. . E x p r e s s i o n .2$ $.. color(red)(Expression.2)$ into groups:

$\Rightarrow (v^{2} - 96 v) + (1 v - 96)$ $color(blue)(rArr (v^2 - 96v) + (1v - 96))$

Factor out $v$ $color(green)(v)$ from $(v^{2} - 96 v)$ $color(blue)((v^2 - 96v)$ to obtain $\Rightarrow v \cdot (v - 96)$ $color(blue)(rArr v*(v - 96) )$

Factor out $1$ $color(green)(1)$ from $(1 v - 96)$ $color(blue)((1v - 96)$ to obtain $\Rightarrow 1 \cdot (v - 96)$ $color(blue)(rArr 1*(v - 96) )$

$S t e p .4$ $color(green)(Step.4)$

Using $S t e p .3$ $color(green)(Step.3)$ we can factor out the common term $(v + 1)$ $color(blue)((v+1)$ and write the factors of our quadratic expression:

$\Rightarrow (v + 1) \cdot (v - 96)$ $color(blue)(rArr (v + 1) * (v - 96)$

$S t e p .5$ $color(green)(Step.5)$

Now, we are in a position to consider

$v^{2} - 95 v - 96 = 0$ $color(blue)(v^2-95v-96 = 0)$ $. . E q u a t i o n .1$ $.. color(red)(Equation.1)$

We can use the factors and write it as:

$\Rightarrow (v + 1) \cdot (v - 96) = 0$ $color(blue)(rArr (v + 1) * (v - 96) = 0$

$\Rightarrow (v + 1) = 0 or (v - 96) = 0$ $color(blue)(rArr (v + 1) = 0 or (v - 96) = 0$

$\Rightarrow (v = - 1) or (v = 96)$ $color(blue)(rArr (v = - 1) or (v = 96)$

So our final solution set is given by

$\Rightarrow v = - 1, v = 96$ $color(blue)(rArr v = - 1, v = 96$

I hope this helps.

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How do you solve $v^{2} - 95 v - 96 = 0$ $v^ { 2} - 95v - 96= 0$ ?

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How do you solve $v^{2} - 95 v - 96 = 0$ $v^ { 2} - 95v - 96= 0$ ?