Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship betwee
the zeroes and the coefficients.
(1) x2 - 2x -8
(ii) 4.s2 - 4s +1
(i) 6x2 -3-7x
((iv) 4u² + 8u
(v) - 15
(vi) 3x2 - x-4
2. Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively.
(m) 0,75
(1) -:-1
4
-
(vi) 4,1
(iv) 1,1
1 1
4' 49
(v).
please solve this both questions. ​

Answer 1

Step-by-step explanation:

I solve 1 example of each question

1.

${x}^{2} - 2x - 8 = 0 \\ by \: frctorisation \\ {x}^{2} -4x + 2x - 8 = 0 \\ x(x - 4) + 2(x - 4) = 0 \\ (x + 2)(x - 4) = 0 \\ x + 2 = 0 \: \: or \: \: x - 4 = 0 \\ x = - 2 \: \: orx = 4$

We find the factors of quadratic polynomial

${x}^{2} - 2x - 8$

are 2 & - 4

When the coefficient of x^2 is 1

Then the zeros are opposite number of factors

2.

vi) 4,1

$let \: \: \alpha = 4 \: \: and \: \beta = 1 \\ \alpha + \beta = 4 + 1 = 5 \\ \alpha \times \beta = 4 \times 1 = 4 \\ the \: required \: quadratic \: eqution \: is \: \\ {x}^{2} - ( \alpha + \beta )x + \alpha \times \beta = 0 \\ {x}^{2} - 5x + 4 = 0$

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