The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
Answer:
No
Step-by-step explanation:
We have to find that a quadratic polynomial equation with real coefficient can have one real solution and one complex solution.
quadratic equation is given by
It can be written as the product of linear factors
Where are solutions of the given polynomial equation.
No , a quadratic polynomial equation can not have one real solution and one complex solution because complex root are always in paired not a single.
A quadratic equation have two roots only.
If a quadratic equation have complex root then the equation have both complex root .
If a equation have real root then it have both real.
Therefore, a quadratic equation can not have one real and one compelx solution.