Question

Which among the following values of x $x=9,\dfrac{1}{2},3$ satisfy the equation $3{{x}^{2}}-8x-3=0$

Answer 1

Hint: To find if the given values of ‘x’ satisfy the given equation or not, substitute the values of ‘x’ in the quadratic equation and simplify them. Another way to check if the values of ‘x’ satisfy the given equations or not is by factorizing the equation by splitting the middle term or completing the square method.

Complete step-by-step answer:
We have to check if $x=9,\dfrac{1}{2},3$ satisfy the equation $3{{x}^{2}}-8x-3=0$ or not.
We observe that the equation $3{{x}^{2}}-8x-3=0$ is a quadratic equation. The degree of a polynomial is the largest exponent of one of the variables. Thus, the degree of the given quadratic equation $3{{x}^{2}}-8x-3=0$ is 2.
We will substitute the given values of ‘x’ in the equation and check if they satisfy the equation or not.
Substituting $x=9$ in the equation $3{{x}^{2}}-8x-3=0$, we have $3{{\left( 9 \right)}^{2}}-8\left( 9 \right)-3=243-72-3=168$. We observe that $168\ne 0$. Thus $x=9$ doesn’t satisfy the given equation.
We will now substitute $x=\dfrac{1}{2}$ in the equation $3{{x}^{2}}-8x-3=0$. Substituting the value, we have $3{{\left( \dfrac{1}{2} \right)}^{2}}-8\left( \dfrac{1}{2} \right)-3=\dfrac{3}{4}-4-3=\dfrac{3}{4}-7=\dfrac{-25}{4}$. We observe that $\dfrac{-25}{4}\ne 0$. Thus $x=\dfrac{1}{2}$ doesn’t satisfy the given equation.
We will now substitute $x=3$ in the equation $3{{x}^{2}}-8x-3=0$. Substituting the value, we have $3{{\left( 3 \right)}^{2}}-8\left( 3 \right)-3=27-24-3=0$. We observe that $0=0$. Thus $x=3$ satisfy the given equation.
Hence, $x=3$ satisfies the equation $3{{x}^{2}}-8x-3=0$.

Note: We can also solve this equation by factorizing the equation $3{{x}^{2}}-8x-3=0$ using splitting the middle term method. We can rewrite the given equation as $3{{x}^{2}}-9x+x-3=0$. Taking out the common terms, we have $3x\left( x-3 \right)+1\left( x-3 \right)=0$. Further simplifying the previous equation, we have $\left( x-3 \right)\left( 3x+1 \right)=0$. Thus, we have $x=3,\dfrac{-1}{3}$. We can also factorize the given equation by completing the square method or using algebraic identities.