Question

How do you use the discriminant to find all the values of c for which the equation \[2{{x}^{2}}+5x+c=0\] has two real roots?

Answer 1

Hint: We are given an equation as \[2{{x}^{2}}+5x+c=0\] and we are asked to find the value of c for which we can have real roots. To find this, we will learn how to find the discriminant firstly and then we will learn how the value of the discriminant responds to the behavior of the roots. Then we will use \[2{{x}^{2}}+5x+c=0\] and find the discriminant of it. We will use that discriminant is given as \[\Delta ={{b}^{2}}-4ac.\]

Complete step by step answer:
We are given an equation as \[2{{x}^{2}}+5x+c=0\] and we have to find those c from which this equation has two real roots. We have to decide this using the value of the discriminant. To answer this we will learn what does the discriminant means and how its value affects the roots. Now, we know that the discriminant of a polynomial is a quantity that depends on the coefficient and determines the various properties of the root. As we can see that our polynomial \[2{{x}^{2}}+5x+c=0\] has degree 2. So, the discriminant of the quadratic polynomial \[a{{x}^{2}}+bx+c=0\] with \[a\ne 0\] is given as \[\Delta ={{b}^{2}}-4ac.\]
Now, for different values of the discriminant, the root has different behavior. If the discriminant is 0 then the quadratic equation has real and equal roots. If the discriminant is positive \[\left( \Delta >0 \right)\] then the quadratic equation has real and distinct roots and lastly, if the discriminant is negative \[\left( \Delta <0 \right)\] then the quadratic equation has complex roots. As we are asked that our equation must have a real root. So it may be equal or distinct means the value of the discriminant for \[2{{x}^{2}}+5x+c=0\] is greater than or equal to zero, then this equation has real roots. So for \[2{{x}^{2}}+5x+c=0\] we have a = 2, b = 5 and c = c. So as \[\Delta \ge 0,\] so \[{{b}^{2}}-4ac\ge 0.\] So, using the above values we get,
\[{{5}^{2}}-4\times 2\times c\ge 0\]
\[\Rightarrow 25-8c\ge 0\]
Adding 8c to both sides, we get,
\[\Rightarrow 25\ge 8c\]
Dividing both sides by 8, we get,
\[\Rightarrow \dfrac{25}{8}\ge c\]
So, we get,
\[\Rightarrow c\le \dfrac{25}{8}\]
If c is less than or equal to \[\dfrac{25}{8}\] then the roots are always real.

Note: While solving the inequality \[\left( <,>,\le ,\ge \right)\] we need to understand that when we multiply the inequality by a negative term, the inequality always gets reversed. If we multiply the inequality by a positive term, then there is no effect on the inequality. We need to be careful that when we decide the value of a, b and c from \[a{{x}^{2}}+bx+c=0,\] we pick the value along with their sign. Mistakes happen like \[2{{x}^{2}}-4x+6\] here a = 2, b = – 4 and c = 6. One may write a = 2, b = 4 and c = 6. So, it can cause problems.