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Question: How do you solve \(2{x^2} - 3x - 9 = 0\)?...

How do you solve 2x2−3x−9=02{x^2} - 3x - 9 = 0?

Explanation

Solution

There are various methods by which we can solve the given equation i.e., factorization, completing the square or quadratic formula. Let us solve the given equation for the value of xx by quadratic formula. Before solving this question, we will first have to compare the given equation with the standard quadratic equation, which is ax2+bx+c=0a{x^2} + bx + c = 0, wherea≠0a \ne 0. After comparing both the equations with each other, we will have to find the values of a,ba,b and cc. Next, we will have to substitute the obtained values in the quadratic formula.
Formula used:
Quadratic formula: −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step solution:
The quadratic equation is 2x2−3x−9=02{x^2} - 3x - 9 = 0.
We have to solve the given equation for finding out the value of xx. Before starting with the solution, we first need to compare the given equation with the standard quadratic equation.
Standard quadratic equation: ax2+bx+c=0a{x^2} + bx + c = 0.
Now, let 2x2−3x−9=02{x^2} - 3x - 9 = 0-----(1)
After comparing equation (1) with the standard quadratic equation we get, a=2,b=−3a = 2,b = - 3 and c=−9c = - 9
As for the next step, we have to substitute the obtained values of a,ba,b and cc in the quadratic formula in order to find the values of xx.
After substituting the values, we get:
⇒−(−3)±(−3)2−4(2)(−9)2(2) ⇒3±9+724 ⇒3±814 ⇒3±94 ∴x=3+94,3−94 x=124,−64 x=3,−32  \Rightarrow \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 2 \right)\left( { - 9} \right)} }}{{2\left( 2 \right)}} \\\ \Rightarrow \dfrac{{3 \pm \sqrt {9 + 72} }}{4} \\\ \Rightarrow \dfrac{{3 \pm \sqrt {81} }}{4} \\\ \Rightarrow \dfrac{{3 \pm 9}}{4} \\\ \therefore x = \dfrac{{3 + 9}}{4},\dfrac{{3 - 9}}{4} \\\ x = \dfrac{{12}}{4}, - \dfrac{6}{4} \\\ x = 3, - \dfrac{3}{2} \\\
Thus, we can say that x=3x = 3 or x=−32x = - \dfrac{3}{2}.
Hence, the quadratic equation 2x2−3x−9=02{x^2} - 3x - 9 = 0 after solving with quadratic formula has roots x=3x = 3orx=−32x = - \dfrac{3}{2}.

Note: In the quadratic formula −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, the expression b2−4ac\sqrt {{b^2} - 4ac} is called as discriminant and is often denoted by DD. Now, depending upon the quadratic equation, the value of discriminant changes and therefore, the value of roots also change. If DD is positive or greater than zero, then the two roots of the equation are real. If DD is zero, then roots are real but if DD is negative or less than zero, then roots are not real.