Question

Solve the following quadratic equation for: \[x:4\sqrt{3}{{x}^{2}}+5x-2\sqrt{3}=0\].

Answer 1

- Hint: First get a quadratic expression from the given equation. Now take the 3 coefficients and mark them as a, b, c. Find value of product ac. Now, try to write b as the sum or difference of 2 numbers which when multiplied you must get ac. Now take the common from the first two terms and last two terms. Such that there must by common terms in both the above terms. Basically, we are converting the quadratic into multiplication of 2 uni degree terms. That is when we multiply these 2 1 degree terms we must get the given quadratic.


Complete step-by-step solution -

Factorization: For factorizing a quadratic follow the steps:
Allot ${{x}^{2}}$ coefficient as “a”, x co-efficient as “b”, constant as “c”.
Find the product of the 2 numbers a, c. Let it be k.
Find 2 numbers such that their product is k, sum is b.
Rewrite the term $bx$ in terms of those 2 numbers.
Now, factor the first two terms.
Now, factor the last two terms.
If we do it correctly our 2 new terms will have one more common factor.
Now take that common to get quadratic as a product of 2 terms, from which you get the roots.

Given expression in the question can be written as:
\[4\sqrt{3}{{x}^{2}}+5x-2\sqrt{3}=0\]
Here, we get \[a=4\sqrt{3},b=5,c=-2\sqrt{3}\] . So, product ac written as:
\[\Rightarrow a\cdot c=4\sqrt{3}\cdot \left( -2\sqrt{3} \right)=-24\]
So, the 2 numbers whose product is \[-24\] and sum is 5 are:
\[8,-3\] \[\left( \because 8+\left( -3 \right)=5;\left( 8 \right)\cdot \left( -3 \right)=-24 \right)\]
By this we can write our quadratic equation as:
\[\Rightarrow 4\sqrt{3}{{x}^{2}}+8x-3x-2\sqrt{3}=0\]
Now take 4x common from first two terms, we get:
\[\Rightarrow 4x\left( \sqrt{3}x+2 \right)-3x-2\sqrt{3}=0\]
Now take \[\left( \sqrt{3}x+2 \right)\] common from whole equation, we get:
\[\Rightarrow \left( \sqrt{3}x+2 \right)\left( 4x-\sqrt{3} \right)=0\]
By this, we get two equation which can be written as:
\[\sqrt{3}x+2=0\] ; \[4x-\sqrt{3}=0\]
From these we get the roots as given below:
\[x=-\dfrac{2}{\sqrt{3}}\] ; \[x=\dfrac{\sqrt{3}}{4}\]
Therefore, \[-\dfrac{2}{\sqrt{3}},\dfrac{\sqrt{3}}{4}\] are roots of given quadratic equations.

Note: The idea of getting 2 numbers fitting the required condition is the crucial part, so do it properly. Instead of writing \[8x-3x\] you can also write \[-3x+8x\] , then you must take \[\sqrt{3}x\] as common from the first two terms and 2 common from last two terms. Even then, you will reach the same result. So, it doesn’t matter in which order you use the 2 numbers.