Question

How do you solve the equation ${{v}^{2}}+2v-8=0$ .

Answer 1

Hint: Now to solve the quadratic equation we will use splitting the middle term method. In this we will split the middle terms such that their product gives the product of the other two terms. Now to the obtained equation we will use the distributive property to simplify the equation. Hence we will get the factors of the equation by which we can easily find the roots of the given equation.

Complete step-by-step solution:
Now consider the given equation ${{v}^{2}}+2v-8=0$ .
Now we want to find the solution of the equation. Which means we want to find the roots of the equation.
Roots of an equation means the values at which the equation is satisfied. Hence we want to find v such that the equation given will be satisfied.
Now to do so we will first split the middle term of the equation such that the addition of the terms gives the middle term and the multiplication of the coefficient of the terms give us constant × coefficient of ${{x}^{2}}$.
For example here we split 2v and 4v – 2v as 4 × (- 2) = - 8 = - 8 × 1.
Hence we split the middle terms and rewrite the equation as ${{v}^{2}}+4v-2v-8=0$
Now taking v common from the first two terms and -2 common from last two terms we get,
$\Rightarrow v\left( v+4 \right)-2\left( v+4 \right)=0$
Now taking the term v + 4 common we get,
$\Rightarrow \left( v+4 \right)\left( v-2 \right)=0$
Now if multiplication of two terms is 0 then one of the terms must be zero.
Hence v = -4 or v = 2.
Hence the solution of the equation is v = - 4 or v = 2.

Note: Now note that to find the roots of a quadratic equation we can use different methods. We can also find the roots by a complete square method in which we add and subtract certain terms such that we get a perfect square on one side of the equation. The roots of the equation of the form $a{{x}^{2}}+bx+c=0$ is can also be given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .