Question

How do you solve $5{{x}^{2}}-2x-6=-3{{x}^{2}}+6x$ ?

Answer 1

Hint: In this question, we have to find the value of x. So, to solve this problem, we have to rewrite the equation in the general form of the quadratic equation $a{{x}^{2}}+bx+c=0$ and then we will apply splitting the middle method, to get two values of x. First, we will add $3{{x}^{2}}$ on both sides of the equation, then subtract 6x and make necessary calculations to get the new equation in the general form of the quadratic equation. After that, we will split the middle term of the equation in the sum of $4x$ and $-12x$ . After that, we will take common 4x in the first two terms and -6 in the last two terms. After the necessary calculations, we get two values of x, which is our required solution.

Complete step-by-step answer:
According to the question, we have to find the value of x.
Thus, to solve this problem we will use the splitting the middle terms method.
The equation given to us is $5{{x}^{2}}-2x-6=-3{{x}^{2}}+6x$ ---------- (1)
First, we will rewrite the equation (1) in the general form of a quadratic equation $a{{x}^{2}}+bx+c=0$, that is
We will add $3{{x}^{2}}$ on both sides in the equation (1), we get
$\Rightarrow 5{{x}^{2}}-2x-6+3{{x}^{2}}=-3{{x}^{2}}+6x+3{{x}^{2}}$
As we know, the same terms with opposite signs cancel out each other, therefore we get
$\Rightarrow 5{{x}^{2}}-2x-6+3{{x}^{2}}=6x$
On further simplification, we get
$\Rightarrow 8{{x}^{2}}-2x-6=6x$
Now, subtract 6x on both sides of the above equation, we get
$\Rightarrow 8{{x}^{2}}-2x-6-6x=6x-6x$
As we know, the same terms with opposite signs cancel out each other, therefore in RHS we get
$\Rightarrow 8{{x}^{2}}-2x-6-6x=0$
On further solving, we get
$\Rightarrow 8{{x}^{2}}-8x-6=0$ -------------- (2)
As we see know, that equation (2) is in the general form of the quadratic equation $a{{x}^{2}}+bx+c=0$ , thus on comparing both the equations, we get
$a=8\text{, }b=-8\text{, and }c=-6$
Therefore, we will apply splitting the middle term method, that is, we will express the middle term of equation $b$ in such a way that it will be the sum of the factors of $a.c$ .
So, we see that $ac=8.(-6)=-48$ , that is
$-48=(+4).(-12)$ and $+4-12=-8$
So, we will rewrite the middle term as a sum of 4x and -12x, we get
$\Rightarrow 8{{x}^{2}}+4x-12x-6=0$
Therefore, we take 4x common in the first two terms and -6 common in the last two terms, we get
 $\Rightarrow 4x(2x+1)-6(2x+1)=0$
Now, we take common (2x+1) in the above equation, we get
$\Rightarrow (2x+1)(4x-6)=0$
So, either $2x+1=0$ ----------- (3) or
$4x-6=0$ --------- (4)
Thus, first, we will solve equation (3), that is
$\Rightarrow 2x+1=0$
We will subtract 1 on both sides in the above equation, we get
 $\Rightarrow 2x++1-1=0-1$
As we know, the same terms with opposite signs cancel out, therefore we get
$\Rightarrow 2x=-1$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}x=\dfrac{-1}{2}$
Therefore, we get
$\to x=-\dfrac{1}{2}$
Now, we will solve equation (4), which is
$4x-6=0$
Now, we will add 6 on both sides in the above equation, we get
$\Rightarrow 4x-6+6=0+6$
As we know, the same terms with opposite signs cancel out, we get
$\Rightarrow 4x=6$
Now, we will divide 4 on both sides of the equation, we get
$\Rightarrow \dfrac{4}{4}x=\dfrac{6}{4}$
$\Rightarrow x=\dfrac{3}{2}$
Therefore, for the equation $5{{x}^{2}}-2x-6=-3{{x}^{2}}+6x$ , the value of x is $-\dfrac{1}{2}$ and $\dfrac{3}{2}$ .

Note: While solving this problem, do mention all the steps properly to avoid confusion and mistakes. One of the alternative methods to solve this problem is to use the discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get the value of x, which is our required solution to the problem.