Question

How do you solve the equation \[4{{x}^{2}}-14x-8=0\]?

Answer 1

Hint: This type of problem is based on the concept of factoring and solving x. First, we have to consider the given equation and divide the whole equation by 2. Then, split the middle term of the equation in such a way that it has a common term in both the first and last term. Here, the middle term is -7x, so we can write -7x as the sum of -8x and x. Then, take the common terms from the obtained equation and represent the equation as a product of two functions of x. Thus, the two functions are the factors of the given equation which on equating to 0, we get the final answer.

Complete step by step solution:
According to the question, we are asked to solve \[4{{x}^{2}}-14x-8=0\].
 We have been given the quadratic equation is \[4{{x}^{2}}-14x-8=0\]. ---------(1)
Let us first divide the whole equation (1) by 2.
\[\dfrac{4{{x}^{2}}-14x-8}{2}=\dfrac{0}{2}\]
On further simplification, we get
\[\dfrac{4{{x}^{2}}}{2}-\dfrac{14x}{2}-\dfrac{8}{2}=\dfrac{0}{2}\]
On cancelling out the common terms, we get
\[2{{x}^{2}}-7x-4=0\] ----------(2)
Let us now split middle term in such a way that the sum of the two terms is equal to -7 and the product of the two terms is equal to -8.
We know that \[-8\times 1=-8\] and -8+1=-7.
Let us substitute in equation (2).
We get
\[2{{x}^{2}}+\left( -8+1 \right)x-4=0\]
On using distributive property in the obtained equation, that is, \[a\left( b+c \right)=ab+ac\]
We get
\[2{{x}^{2}}-8x+1x-4=0\]
Let us now find the common term.
\[2{{x}^{2}}-2\times 4x+1x-4=0\]
Here, 2x is the common term from the first two terms and 1 is common from the last two terms.
Therefore, \[2x\left( x-4 \right)+1\left( x-4 \right)=0\].
From the obtained equation, find that (x-4) is common.
On taking out (x-4) common from the two terms, we get
\[\left( x-4 \right)\left( 2x+1 \right)=0\]
Therefore, the factors are $(x+3) \text{ and } (2x+5)$.
We know that factors are equal to zero if the whole equation is equal to zero.
\[\Rightarrow \left( x-4 \right)=0\] and \[\left( 2x+1 \right)=0\].
Let us now consider $x-4=0.$
Add -4 on both sides. We get
$x-4+4=0+4$
On further simplification, we get
$x=4$
Now, consider $2x+1=0.$
Add -1 on both the sides. We get
$2x+1-1=0-1$
That is, $2x=-1.$
Then, divide the obtained expression by 2. We get
\[x= - \dfrac{1}{2}\]
Hence, the values of x in the given equation \[4{{x}^{2}}-14x-8=0\] are 4 and \[-\dfrac{1}{2}\].

Note: Whenever we get such types of problems, we should make necessary calculations to the given quadratic equation and then take out the common terms out of the bracket to obtain the values of x. Also, avoid calculation mistakes based on the sign conventions. Here, we can solve this question by using quadratic formula, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. In this question $a=4, b=-14 \text{and} c=-8.$ Substitute these values in the given formula and we can find the values of x.