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Answers
Answer:
1) i) Given :
find the zero of the each of the following and verify the relationship between the zeros and their Coefficient 7 x² - 25 x -12
To find :
Find the zeros and verify the relationship between zeros and their Coefficient
Solution :
7x² - 25x -12 = 0
solve this by splitting middle term
= 7x² - 3x + 28x - 12 = 0
= x(7x-3) + 4(7x -3) = 0
= (x+4)(7x-3) = 0
Either
=> x + 4 = 0
=> x = -4
or
=> 7x - 3 = 0
=> 7x = 3
=> x = 3/7
-4 and 3/7 are the zeros of the given polynomial
Verification :
Add both the zeros
Multiply both the zeros
Hence , it is verified
iii) P(x) = √3x2+10x+7√3
= √3x+3x+7x+7√3
= (√3x+7)(x+√3)
= √3x+7 = 0 and x+√3 = 0
= x = -7/√3 and x = -√3
verification by α and β,
let α = -7/√3 and β = -√3
sum of zeros = α+β = -7/√3+(-√3)
= -7/√3-√3
= -7-3/√3
= -10/√3
product of zeroes = αβ
= -7/√3 . -√3
= 7.
verification by coefficients,
sum of zeros = -b/a
= -10/√3.
product of zeros ,
= c/a
= 7√3/√3
= 7.
hence relationship is verified.
3) Given,
y=x
2
−x−6
x
2
−x−6=0
→(x−3)(x+2)=0
→x=−2,3
y=6−x−x
2
6−x−x
2
=0
→(x+3)(x−2)=0
→x=−3,2
figure is in attachment
4) The polynomial x
3 +2x
2
−x−2=0 can be factorised as follows:
x
3
+2x
2
−x−2=0
⟹x
2
(x+2)−1(x+2)=0
⟹(x
2
−1)(x+2)=0
⟹ x
2
−1=0 or x+2=0
⟹ x
2
=1 or x=−2
x=1, x=−1 and x=−2
Hence, the zeroes of the polynomial x
3
+2x
2
−x−2=0 are 1,−1,−2.
6)Given zeroes,
5+√2 and 5-√2
Sum of the zeroes=5+√2+5-√2
Sum of the zeroes=10
Product of zeroes=(5+√2)(5-√2)
Product of zeroes=25-2=23
We know that quadratic polynomial is in the form of,
=k{x²-(sum of zeroes)x+product of zeroes}
By putting required values we get,
=k{x²-10x+23}
=x²-10x+23
Hence x²-10x+23 is the required polynomial.
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