the quadratic equation 2x²– 8x + p = 0 has two distinct real roots. what is the highest possible integer value of p?
Answers
Question :-
The quadratic equation 2x²– 8x + p = 0 has two distinct real roots. What is the highest possible integer value of p ?
Answer :-
Step by step Explanation :-
We Know that If the roots are real & unequal of any quadratic equation ax² + bx + c then its discriminant is greater than zero i.e.
Here We Have,
2x²– 8x + p = 0 has two distinct real roots
Therefore,
So largest possible integer value of p less than 8 will be 7 .
Therefore,
Additional Information :-
❒ Quadratic Polynomial with one Variable :
✪ The general form of the equation is ax² + bx + c = 0.
Note :
◆ If a = 0, then the equation becomes to a linear equation.
◆ If b = 0, then the roots of the equation becomes equal but opposite in sign.
◆ If c = 0, then one of the roots is zero. ]
❒ Nature Of Roots :
✪ b² - 4ac is the discriminant of the equation.
Then,
● If b² - 4ac = 0, then the roots are real & equal.
● If b² - 4ac > 0, then the roots are real & unequal.
● If b² - 4ac < 0, then the roots are imaginary.
Step-by-step explanation:
Given :
To find :
Solution :
- If discriminant (b² - 4ac) > 0, the quadratic equation has two distinct real roots.
- b² - 4ac > 0
⇒ (-8)² - 4(2)(p) > 0
⇒ 64 - 8p > 0
⇒ 64/8 - 8p/8 > 0
⇒ 8 - p > 0
⇒ -p > -8
- [When dividing with (-) sign, always inequality reverses]