Question

How do you solve $3{{x}^{2}}-10x-8=0$ ?

Answer 1

Hint: The equation given in the question is a quadratic equation , we know the formula for roots of quadratic equation is equal to $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term in a quadratic equation. We will get real roots if the discriminant value which is ${{b}^{2}}-4ac$ is greater than 0.

Complete step-by-step answer:
The given equation in the question is $3{{x}^{2}}-10x-8=0$ which is quadratic equation
If we compare the equation $3{{x}^{2}}-10x-8=0$ to $a{{x}^{2}}+bx+c=0$ the value of a is 3, the value of b is -10 and the value of c is equal to -8
We know the formula for roots of a quadratic equation which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We put the value of a, b, and c in the above formula we can get the value of roots
So the roots are $\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\times 3\times -8}}{2\times 3}$
Further solving we get
$\Rightarrow \dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\times 3\times -8}}{2\times 3}=\dfrac{10\pm 14}{6}$
So the roots are 4 and $-\dfrac{2}{3}$

Note: We can solve the quadratic equation in factorization method. In this method we can write the quadratic equation $a{{x}^{2}}+bx+c=0$ as $a\left( x-\alpha \right)\left( x-\beta \right)=0$ where $\alpha $ and $\beta $ are roots of the equation. To factorize we have to write the term bx as mx+nx such that product of m and n is equal to product of a and c. Let’s solve $3{{x}^{2}}-10x-8=0$ by method of factorization.
We can write $3{{x}^{2}}-10x-8=0$ = $3{{x}^{2}}-12x+2x-8=0$
We can take 3x common in the first half and 2 common in the second half of the equation
$\Rightarrow 3{{x}^{2}}-10x-8=3x\left( x-4 \right)+2\left( x-4 \right)$
We can take x- 4 common from the whole equation
$\Rightarrow 3{{x}^{2}}-10x-8=\left( x-4 \right)\left( 3x+2 \right)$
If $\left( 2x-1 \right)\left( x-3 \right)$ is equal to 0 then x can be 4 or $-\dfrac{2}{3}$
So the roots of the equation $3{{x}^{2}}-10x-8=0$ are 4 and $-\dfrac{2}{3}$