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If α, β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7. Find a polynomial whose zeros are 2α + 3β and 3α + 2β.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
If α, β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7, the polynomial whose zeros are 2α + 3β and 3α + 2β is 2×2 – 25 + 82. Candidates should write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations’ answers are the solutions to the derived equations. In this post, we will go through the step-by-step solution for the question “If α, β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7. Find a polynomial whose zeros are 2α + 3β and 3α + 2β.”
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f(x) = 2×2 – 5x + 7. Find a polynomial whose zeros are 2α + 3β and 3α + 2β.
The question states If α, β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7. Find a polynomial whose zeros are 2α + 3β and 3α + 2β. At most, a polynomial function can have real roots equal to its degree. Set a function’s value to zero and solve to find its roots. The steps to find a polynomial whose zeros are 2α + 3β and 3α + 2β are as follows:
Given that p(x) = 2×2 – 5x + 7 and its zeros are denoted by α, β
Now we have to compare p(x) = 2×2 – 5x + 7 with ax2 + bx + c
a = 2, b = -5 and c =7
we know that, sum of zeros = α + β = -b/a
Substituting the values we get,
= 5/2
product of zeros = c/a
Substituting the values we get:
= 7/2
2α + 3β and 3α + 2β are the zeros of the required polynomial,
sum of zeros = 2α + 3β + 3α + 2β
= 5α + 5β
Taking common,
= 5 (α + β)
= 5 x 5/2
Substituting the values we get:
= 25/2
product of zeros = (2α + 3β) (3α + 2β)
= 2α (3α + 2β) + 3β (3α + 2β)
On multiplying the above equation we get:
= 6α2 + 4αβ + 9αβ + 6β2
On simplifying:
= 6α2 + 13αβ + 6β2
= 6 [α2 + β2] + 13αβ
= 6 [(α + β)2 – 2αβ] + 13αβ
Substituting the values we get:
= 6 [(5/2)2 – 2 x 7/2] + 13 x 7/2
= 6 [25/4 – 7] + 91/2
= 6 [25 – 28/4] + 91/2
= 6 [-¾] + 91/2
= -18/4 + 91/2
= -9/2 + 91/2
= 82/2
= 41
A quadratic polynomial is given by:
k [x2 – (sum of zeros)x + (product of zeros)]
= k [x2 – 25x/2 + 41]
Take k = 2
= 2 [x2 – 25x/2 + 41]
= 2×2 – 25 + 82 is the required polynomial.
Summary:
If α, β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7. Find a polynomial whose zeros are 2α + 3β and 3α + 2β.
If α and β are the zeros of the quadratic polynomial f(x) = 2×2 – 5x + 7. As we all know that the sum of zeros = α + β = -b/a and the product of zeros = c/a. A quadratic polynomial is given by k [x2 – (sum of zeros)x + (product of zeros)]. By substituting all values we will get the desired polynomial whose zeros are 2α + 3β and 3α + 2β. The polynomial whose zeros are 2α + 3β and 3α + 2β is 2×2 – 25 + 82.
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