Question

How do you solve \[12{{x}^{2}}-25x+12>=0\] ?

Answer 1

Hint: We will solve this inequality the same as we solve our linear equation. We solve this question using the factorization method. We will split the middle term in a way that their sum is equal to the middle term and product is equal to product of first and last terms. Finally, we have to take terms as common and simplify them. we have to represent the equation as a product of \[\left( x-a \right)\left( x-b \right)\].

Complete step by step solution:
Here we will factorize the quadratic polynomial \[12{{x}^{2}}-25x+12>=0\].
Now we have to split the middle term into two in such a way that the sum is equal to \[-25\] and the product is equal to the product of constant term \[12\]and \[12{{x}^{2}}\]that is \[144\].
To do this, first we need to prime factorize for the number \[144\]
So let us find prime factors for \[144\]
We can write \[144\] as
 \[144=2\times 2\times 2\times 2\times 3\times 3\] as a product of primes.
Now we can take -16 and \[-9\] to satisfy the given condition.
\[\begin{align}
  & 144=-16\times -9 \\
 & -25=-9-16 \\
\end{align}\]
So these values satisfy the condition now we will split the middle term as \[-16x-9x\]
Then the polynomial will look like
\[\Rightarrow 12{{x}^{2}}-16x-9x+12\]
Now we can take common terms out to make them as factors.
Here we can see from the first two terms we can take \[4x\] as common and from the next two terms we can \[6\] as common . After taking the common terms out the equation will look like
\[\Rightarrow 4x\left( 3x-4 \right)-3\left( 3x-4 \right)\]
Now in the above polynomial we can see \[4x-3\] as common and we can take it out then the polynomial will look like
\[\Rightarrow \left( 3x-4 \right)\left( 4x-3 \right)\]
Now we can solve these factors for getting values also.
\[\left( 3x-4 \right)\left( 4x-3 \right)>=0\]
We can write these as
\[\Rightarrow 3x-4>=0\]
\[\Rightarrow 4x-3>=0\]
We will get
\[\Rightarrow 3x>=4\]
\[\Rightarrow 4x>=3\]
By simplifying these we will get
\[\Rightarrow x>=\dfrac{4}{3}\]
\[\Rightarrow x>=\dfrac{3}{4}\]
So the values of \[x\] are greater than or equal to \[\dfrac{4}{3}\] and \[\dfrac{3}{4}\].

Note:
We can also solve the above question using discriminant methods also. In which we can solve the x for values a and b after that making them factors as \[\left( x-a \right)\left( x-b \right)\]. Here we will find factors. First we will find values of \[x\].