Question

How do you solve $3{x^2} - 5x + 7 = 0$ ?

Answer 1

Hint:Firstly, the given equation is quadratic. The solution of the quadratic equation will be the roots of the equation. The roots of the equation can be found out by using the roots for a quadratic equation formula. We get two values from the formula both of which are the roots of the equation.
Formula used:
The roots of a quadratic equation( $a{x^2} + bx + c = 0$ ) can be found by the formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$

Complete step by step answer:The given expression is, $3{x^2} - 5x + 7 = 0$
Now use the formula for finding the roots of the above quadratic expression.
The formula is $x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$
Here, $a = 3;b = ( - 5);c = 7$
Since the value of
${b^2} - 4ac = {( - 5)^2} - 4(3 \times 7) = - 59$
Which is $< 0$ . This means that our roots are going to be imaginary and complex.
The solution which we are going to get is not real.
On substituting these values in the formula, we get,
$\Rightarrow x = \left[ {\dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \times 3 \times 7} }}{{2 \times 3}}} \right]$
First, solve the operations in the square root.
$\Rightarrow x = \left[ {\dfrac{{5 \pm \sqrt {25 - 84} }}{{2 \times 3}}} \right]$
$\Rightarrow x = \left[ {\dfrac{{5 \pm \sqrt { - 59} }}{{2 \times 3}}} \right]$
Write the values of the square root.
$\Rightarrow x = \left[ {\dfrac{{5 \pm \sqrt { - 59} }}{6}} \right]$
Open the $\pm$ sign to get two values of $x$
$\Rightarrow x = \left[ {\dfrac{{5 + \sqrt { - 59} }}{6}} \right];x = \left[ {\dfrac{{5 - \sqrt { - 59} }}{6}} \right]$The discriminant is negative, so the solutions are not real.
Since we can’t evaluate any further, we write to them as it is.
$\therefore$Hence the roots of the expression, $3{x^2} - 5x + 7 = 0$ are $x = \left[ {\dfrac{{5 + \sqrt { - 59} }}{6}} \right],\left[ {\dfrac{{5 - \sqrt { - 59} }}{6}} \right]$ and are not real.

Additional information: A polynomial is a mathematical expression that contains one or more variables in sum or subtraction format with different powers. Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial.

Note:
Never forget to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with $+$ and another time with $-$ . You can always cross-check your answer by placing the value of $x$ back in the equation. If you get LHS = RHS then your answer is correct.