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Question: Solve the expression \(5 - 4x(1 + 3x)\) \(A)(1 + 2x)(5 + 6x)\) \(B)(1 - 2x)(5 + 6x)\) \(C)(1 -...

Solve the expression 54x(1+3x)5 - 4x(1 + 3x)
A)(1+2x)(5+6x)A)(1 + 2x)(5 + 6x)
B)(12x)(5+6x)B)(1 - 2x)(5 + 6x)
C)(12x)(56x)C)(1 - 2x)(5 - 6x)
D)(1+2x)(56x)D)(1 + 2x)(5 - 6x)

Explanation

Solution

Here we are asked to solve the given quadratic equation. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.
The word quadratic means second-degree values of the given variables.
Then by further simplifying, the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 can be written in the form of (xα)(xβ)=0(x - \alpha )(x - \beta ) = 0.
In this equation, the factors of the quadratic equation are (xα)(x - \alpha ) and (xβ)(x - \beta ).
So, to determine the roots of the equation, these factors of the equation will be made equal to 00.
That is,
(xα)=0(x - \alpha ) = 0 (xβ)=0(x - \beta ) = 0
Which gives,
x=αx = \alpha x=βx = \beta
So, α\alpha and β\beta are the roots of the quadratic equation.

Complete step-by-step solution:
Since given that 54x(1+3x)5 - 4x(1 + 3x) and by the multiplication operation we get 54x(1+3x)=54x12x25 - 4x(1 + 3x) = 5 - 4x - 12{x^2}
Since given that 12x24x+5=0 - 12{x^2} - 4x + 5 = 0. Now convert the value 4x=6x10x - 4x = 6x - 10x then we get 12x2+6x10x+5=0 - 12{x^2} + 6x - 10x + 5 = 0
Now taking the common values out then we have
12x2+6x10x+5=0- 12{x^2} + 6x - 10x + 5 = 0
6x(2x+1)+5(2x+1)=0\Rightarrow 6x( - 2x + 1) + 5( - 2x + 1) = 0
Then by further simplifying, the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 can be written in the form of (xα)(xβ)=0(x - \alpha )(x - \beta ) = 0. Hence, we get
6x(2x+1)+5(2x+1)=06x( - 2x + 1) + 5( - 2x + 1) = 0
(12x)(6x+5)=0\Rightarrow (1 - 2x)(6x + 5) = 0
Therefore 54x(1+3x)5 - 4x(1 + 3x) can be simplified into (12x)(6x+5)=0(1 - 2x)(6x + 5) = 0
Thus, the option B)(12x)(5+6x)B)(1 - 2x)(5 + 6x) is correct.

Note: The quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 can be factored in the form of (xα)(xβ)=0(x - \alpha )(x - \beta ) = 0 where (xα)(x - \alpha ) and (xβ)(x - \beta ) are the factors of the quadratic equation so α\alpha and β\beta are the roots of the quadratic equation and the values of aa cannot be 00.
If the value of aa is 00, then the quadratic equation will become a binomial equation.
Using the multiplication operation we found the values 54x(1+3x)=54x12x25 - 4x(1 + 3x) = 5 - 4x - 12{x^2}
We are also able to make use of the formula of the quadratic equation. Let ax2+bx+c=0a{x^2} + bx + c = 0 be a quadratic equation then the roots of this equation are given by b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.