The roots of the equation x2 + (2p-1)x + p2 = 0 are real if.
a) p 21
b) p < 4
c) p > 1/4
d) p < 1/4
Answers
Step-by-step explanation:
Answer:-
We know that, the roots of a quadratic equation are real only if Discriminant (D) ≥ 0.
given quadratic equation : x² + (2p - 1)x + p² = 0
On comparing it with standard form of a quadratic equation i.e., ax² + bx + c = 0 ;
Let,
a = 1
b = 2p - 1
c = p².
We know,
D or ∆ = b² - 4ac
So,
⟹ b² - 4ac ≥ 0
⟹ (2p - 1)² - 4(1)(p²) ≥ 0
using (a - b)² = a² + b² - 2ab in LHS we get,
⟹ (2p)² + 1² - 2(2p)(1) - 4p² ≥ 0
⟹ 4p² + 1 - 4p - 4p² ≥ 0
⟹ 1 - 4p ≥ 0
⟹ 1 ≥ 4p
⟹ 1/4 ≥ p
∴ The roots of given quadratic equation are real only if p ≤ 1/4.
Answer:
To find : The distance between the school and his house
Solution :
Speed of a student = 2 ½ km/h = 5/2km/h
★ A student reaches his school 6 minutes late.
Consider the time be x
60min = 1 hour
Time = (x + 6) min = (x + 6/60) = (x + 1/10)h
As we know that
★ Speed = distance/time
Consider the distance be y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}⟹25=x+101y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }⟹25=1010x+1y
\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}⟹25=y×10x+110
\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}⟹25=10x+110y
\implies \sf 5(10x +1) = 20y⟹5(10x+1)=20y
\implies \sf 50x +5=20y⟹50x+5=20y
\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)⟹50x−20y=−5(Equation1)
★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}⟹25+1=x−101y
\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}⟹25+2=1010x−1y
\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}⟹27=10x−110y
\implies \sf 7(10x - 1) = 20y⟹7(10x−1)=20y
\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)⟹70x−20y=7(Equation2)
Subtract both the equations
→ 50x - 20y - (70x - 20y) = -5 - 7
→ 50x - 20y - 70x + 20y = - 12
→ - 20x = - 12
→ x = 12/20 = 3/5
Put the value of x in eqⁿ (1)
→ 50x - 20y = - 5
→ 50 × 3/5 - 20y = - 5
→ 30 - 20y = - 5
→ 30 + 5 = 20y
→ 35 = 20y
→ y = 35/20 = 7/4
→ y = 1 3/4 km
•°• The distance between school and house is 1 3/4 km.