Question-1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution: (i) x² - 2x - 8
x² -2x -8=0
x² -(4-2)x-8=0
x² -4x+2x-8=0
(x² -4x)+(2x-8) =0
x(x-4) + 2(x-4) =0
(x+2) (x-4)=0
these are the factors of the equation.But we have to find the value of x.
(x+2) = 0 and (x-4)=0
x = -2 x= 4
Therefore zeroes of polynomial (x²-2x-8) are 4 and -2.
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Now, second part of question is verify the relationship.
Sum of zeroes of given equation =(-2+4) = 2
Now,
Product of zeroes of given equation = (-2 x 4) = -8
Hence the relationship is verified.
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(ii) 4s² - 4s +1
4s² - 4s +1 = 4s² -4s +1
= 4s² -(2+2)s +1
= 4s² -2s-2s +1
= (4s² -2s) - (2s +1)
= 2s(2s-1) -1(2s-1)
= (2s-1) (2s-1)
So, factors of equation are (2s-1), (2s-1)
now find the value of s
2s-1= 0 2s-1= 0
2s= 1 2s=1
s= 1/2 s= 1/2
So, zeroes of polynomial (4s² -4s +1) are 1/2 and 1/2
Now, verification
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Now,
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(iii) 6x² -3 -7x
6x² -7x -3
= 6x² -(9-2)x -3
= 6x² -9x+2x -3
= (6x² -9x)+(2x -3)
= 3x(2x -3) +1(2x -3)
= (3x +1) (2x -3)
Now, find the value of x
3x +1=0 2x -3= 0
3x = -1 2x = 3
x = -1/3 x = 3/2
So,zeroes of polynomial (6x² -7x -3) are -1/3 and 3/2.
Now, verification
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Now,
Hence the relationship is verified.
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(iv) 4u²+8u
4u²+8u = 4u(u+2)
so, here factors of this equation are 4u and (u+2)
Now, we will find the value of u.
4u =0 u+2=0
u=0/4 u= -2
u=0
So, zeroes of polynomial (4u²+8u) are 0 and -2.
Now, verification
Hence the relationship is verified.
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(v) t² - 15
t² - (√15)²
t² - (√15)² = (t-√15)(t+√15)
Now find the value of t
t-√15 = 0 t+√15 = 0
t = √15 t = -√15
So,zeroes of polynomial (t² - 15) are √15 and -√15 .
Now, verification
Sum of zeroes of given equation = -√15 +√15 = 0
Now,
Hence the relationship is verified.
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(vi) 3x² -x -4
3x²-x-4 = 3x² -(4-3)x -4
= 3x² -4x+3x -4
= x(3x-4)+1(3x-4)
= (x+1)(3x-4)
Now find the value of x
x+1 = 0 3x -4 = 0
x = -1 3x = 4
x = 4/3
So,zeroes of polynomial (3x² -x -4) are -1 and 4/3.
Now, verification
Hence the relationship is verified.
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Question-2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Solution: (i) 1/4 , -1
ax²+bx+c=0
First Method
ax²+bx+c=0
Sum of zeroes = -b/a = -1/4
Product of zeroes = c/a = -4/4
means a= 4, b=-1 and c=-4
ax²+bx+c
Quadratic polynomial = 4x²-1x-4 or 4x² -x -4
Second method with formula
Sum of zeroes (𝛂+𝜷) = 1/4
Product of zeroes (𝛂 x 𝜷) = -1
formula for equation = x²-(𝛂+𝜷)x+(𝛂 x 𝜷)
= x²- 1 x +(-1)
4
So, quadratic equation is 4x² -x -4.
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(ii)√2, 1/3
Sum of zeroes (𝛂+𝜷) = √2
Product of zeroes (𝛂 x 𝜷) = 1/3
formula for equation = x² -(𝛂+𝜷)x +(𝛂 x 𝜷)
= x²-(√2)x+ 1/3
So, quadratic equation is 3x² - 3√2x +1
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(iii)0, √5
Sum of zeroes (𝛂+𝜷) = 0
Product of zeroes (𝛂 x 𝜷) = √5
formula for equation = x² -(𝛂+𝜷)x +(𝛂 x 𝜷)
=x² -(0)x+ √5
= x² -0 x x + √5
So, quadratic equation is x² + √5
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(iv) 1, 1
Sum of zeroes (𝛂+𝜷) = 1
Product of zeroes (𝛂 x 𝜷) = 1
formula for equation = x² -(𝛂+𝜷)x +(𝛂 x 𝜷)
=x² -1x+ 1
= x² -x+ 1
So, quadratic equation is x² -x +1
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(v) -1/4, 1/4
Sum of zeroes (𝛂+𝜷) = -1/4
Product of zeroes (𝛂 x 𝜷) = 1/4
formula for equation = x² -(𝛂+𝜷)x +(𝛂 x 𝜷)
So, quadratic equation is 4x² +x +1.
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(vi) 4, 1
Sum of zeroes (𝛂+𝜷) = 4
Product of zeroes (𝛂 x 𝜷) = 1
formula for equation = x² -(𝛂+𝜷)x +(𝛂 x 𝜷)
=x² -4x+ 1
So, quadratic equation is x² -4x +1
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