Question

For the quadratic equation $3{x^2} - 5x + 2 = 0$ . How do you name the larger of the two solutions?

Answer 1

Hint: We have given a quadratic equation of the form $a{x^2} + bx + c = 0$ . To find the roots of the above quadratic equation, we use quadratic formula, which is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Here $a$ is the coefficient of ${x^2}$ , $b$ is the coefficient of $x$ and $c$ is the constant term.

To solve the given quadratic equation, first, we need to determine the values of$a,b$ and $c$. Then we need to substitute the values in the quadratic formula and simplify it.

To name the larger of the two solution first we find the values of $x$ and then we name it.

Complete step-by-step answer:

Step 1: Here the quadratic equation is given as $3{x^2} - 5x + 2 = 0$. In above equation, the coefficient of ${x^2}$ is $3$ so the value of$a = 3$, coefficient of $x$ is $ - 5$ so the value of $b = - 5$ and the constant term is $2$ so the value of $c = 2$ .

Step 2: Now we substitute the values in quadratic formula to solve the given quadratic equation, we get

$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 3 \right)\left( 2 \right)} }}{{2\left( 3 \right)}}$

Step 3: Simplifying the above equation, we get

$x = \dfrac{{5 \pm \sqrt {25 - 24} }}{6}$

Step 4: On further simplification, we get

$x = \dfrac{{5 \pm \sqrt 1 }}{6}$

$x = \dfrac{{5 \pm 1}}{6}$

Step 5: First considering positive sign, we get

$x = \dfrac{{5 + 1}}{6}$

$ \Rightarrow x = 1$

Step 6: Now considering negative sign, we get

$x = \dfrac{{5 - 1}}{6}$

$ \Rightarrow x = \dfrac{2}{3}$

Hence, the solution of the above equation is $x = 1$ and $x =  \dfrac{2}{3}$.

Note:

In the solution of a quadratic equation, there exist two roots.

The discriminant in the quadratic formula is given as $D = {b^2} - 4ac$ .

If the value of discriminant is greater than zero, then we get two different roots, if discriminant is equal to zero, then we get the same roots, and if discriminant is less than zero, then we get imaginary roots.