If a, B are the zeroes of the polynomial 2x²-5x+7, form a quadratic polynomial whose zeroes are 3a and 3B.
Answers
Correct Question
⇒If α, β are the zeroes of the polynomial 2x²+5x-7, form a quadratic polynomial whose zeroes are 3α and 3β .
Solution
Given Equation
⇒2x² +5x - 7 = 0
To Find
⇒α and β
⇒Form a Quadratic equation whose zeroes 3α + 3β
Now Take
⇒2x² +5x - 7 = 0
Use Middle term Splitting
⇒2x² +7x - 2x - 7 = 0
⇒x(2x +7) -1(2x + 7)= 0
⇒(x - 1) = 0 and 2x + 7 = 0
⇒x = 1 and 2x = -7
⇒x = 1 and x = -7/2
We get
⇒α = 1 and β = -7/2
Then
⇒3α = 3 and 3β = -21/2
General Equation
⇒x² - (sum of zeroes)x + Product of zeroes = 0
Now we get
⇒x² - (3-21/2)x + 3×-21/2 = 0
⇒x² - {(6-21)/2}x - 63/2 = 0
⇒x² + (15/2)x - 63/2 = 0
Taking Lcm
⇒2x² + 15x - 63 = 0
Answer
⇒Equation is 2x² + 15x - 63 = 0
Answer:
Correct Question :-
- If α , β are the zeroes of the polynomial 2x² + 5x - 7, form a quadratic polynomial whose are 3α , 3β.
Given :-
- 2x² + 5x - 7
To Find :-
- What is the quadratic polynomial whose zeroes are 3α and 3β.
Solution :-
Given equation :
➲ 2x² + 5x - 7 = 0
↦ 2x² + (7 - 2)x - 7 = 0
↦ 2x² + 7x - 2x - 7 = 0
↦ x(2x + 7) - 1(2x + 7) = 0
↦ (2x + 7)(x - 1) = 0
↦ (2x + 7) = 0
↦ 2x + 7 = 0
↦ 2x = - 7
➦ x = - 7/2
Either,
↦ (x - 1) = 0
↦ x - 1 = 0
➦ x = 1
Hence, we get :
◈ α = 1
◈ β = - 7/2
Now, as we know that :
★ x² - (α + β)x + (αβ) = 0 ★
Then,
➤ 3α = 3 × 1 = 3
➤ 3β = 3 × (- 7/2) = - 7 × 3/2 = - 21/2
Now, according to the question by using the formula we get,
⇒ x² - (3 - 21/2)x + (3 × {- 21/2} ) = 0
⇒ x² - (6 - 21/2)x + (- 63/2) = 0
⇒ x² - (- 15/2)x - 63/2 = 0
⇒ x² + (15/2)x - 63/2 = 0
⇒ 2x² + 15x - 63/2 = 0
By doing cross multiplication we get,
⇒ 2x² + 15x - 63 = 0 × 2
➠ 2x² + 15x - 63 = 0
∴ The quadratic polynomial whose zeroes are 3α and 3β is 2x² + 15x - 63 = 0.