Question
Let p, q 
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
- 15
- 9
- 7
- 8
Hint:
The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations. Here we have to find the number of equation of the form px2 + qx + 1 = 0, having real roots.
The correct answer is: 7
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q {1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q2 ≥ 4
q = 2,3,4
For p=2,q2≥8
q = 3,4
For p=3,q2≥12
q = 4
For p=4,q2≥16
q = 4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so 7 equations can be formed.
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