Question
If
then the equation whose roots are 
- none of these
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 equation whose roots are
.
The correct answer is: 
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 

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, the equation is .
Related Questions to study
Let p, q
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
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.
Let p, q
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
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.
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.
The cartesian equation of
is
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The cartesian equation of
is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is .
The castesian equation of
is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is .
The castesian equation of
is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is .