Question

■▪•°Solve this Equation :-

6x^2 - x - 2 = 0

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Answer 1

6x^2−x−2=0

6x^2−4x+3x−2=0

2x(3x−2)+(3x−2)1=0

(2x+1)(3x−2)=0

2x+1 = 0 or 3x-2 = 0

x = -1/2 , x=2/3.

hope it helps❤

Answer 2

$\Huge\text{ \boxed{x=\dfrac{2}{3}\text{ or }x=\dfrac{1}{2}} }$

$\huge\text{[Topic: Quadratic equations]}$

$\Large\text{Three ways to solve quadratic equations}$

$\large\bullet\text{Factorization}$

→ Factorization solves quadratic equations by finding the value of each factor equals zero.

$\large\bullet\text{Completing the square}$

→ This is also one of the ways to solve quadratic equations, but not commonly. If we solve $\largeax^{2}+bx+c=0\ (a\neq0)$ using this method, we get the quadratic formula.

$\large\bullet\text{Quadratic formula}$

→ The quadratic formula uses the coefficient of each term to find the solutions. It is known by $\large\text{ x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a} }$.

$\huge\text{[Explanation]}$

Let's solve by each method.

$\large\bullet\text{Factorization}$

$\large\cdots\longrightarrow 6x^{2}-x-2=0$

$\large\cdots\longrightarrow (3x+2)(2x-1)=0$

$\large\cdots\longrightarrow 3x+2=0\text{ or }2x-1=0$

$\large\text{ \cdots\longrightarrow\boxed{x=-\dfrac{2}{3}\text{ or }x=\dfrac{1}{2}} }$

$\large\bullet\text{Completing the square}$

All quadratic equations are in the form of $\largeax^{2}+bx+c=0\ (a\neq0)$.

$\large\text{ \cdots\longrightarrow a(x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}})-\dfrac{b^{2}}{4a}+c=0 }$

$\large\text{ \cdots\longrightarrow a(x+\dfrac{b}{2a})^{2}-\dfrac{b^{2}}{4a}+c=0 }$

$\large\text{ \cdots\longrightarrow a(x+\dfrac{b}{2a})^{2}=\dfrac{b^{2}}{4a}-c }$

$\large\text{ \cdots\longrightarrow a(x+\dfrac{b}{2a})^{2}=\dfrac{b^{2}-4ac}{4a} }$

$\large\text{ \cdots\longrightarrow(x+\dfrac{b}{2a})^{2}=\dfrac{b^{2}-4ac}{4a^{2}} }$

$\large\text{ \cdots\longrightarrow x+\dfrac{b}{2a}=\pm\sqrt{\dfrac{b^{2}-4ac}{4a^{2}}} }$

$\large\text{ \cdots\longrightarrow x+\dfrac{b}{2a}=\pm{\dfrac{\sqrt{b^{2}-4ac}}{2a}} }$

$\large\text{ \cdots\longrightarrow\boxed{x={\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}} }$

$\large\bullet\text{Quadratic formula}$

$\begin{cases} &a=6 \\ &b=-1 \\ &c=-2 \end{cases}\rightarrow\boxed{\large{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}$

$\large\text{ \cdots\longrightarrow x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4\cdot6\cdot(-2)}}{2\cdot6} }$

$\large\text{ \cdots\longrightarrow x=\dfrac{1\pm\sqrt{49}}{12} }$

$\large\cdots\longrightarrow x=\dfrac{1\pm7}{12}$

$\large\cdots\longrightarrow x=\dfrac{1+7}{12}\text{ or }x=\dfrac{1-7}{12}$

$\large\cdots\longrightarrow x=\dfrac{8}{12}\text{ or }x=-\dfrac{6}{12}$

$\large\text{ \cdots\longrightarrow\boxed{x=\dfrac{2}{3}\text{ or }x=-\dfrac{1}{2}} }$

In these ways, we solve quadratic equations.