Question

4x(square) + 5 X + 1 Verify relationship with Coefficients .

Answer 1

$4 {x}^{2} + 5x + 1 = 0$

$4 {x}^{2} + 4x + x + 1 = 0$

$4x(x + 1) + 1(x + 1) = 0$

$(x + 1)(4x + 1) = 0$

$x = - 1 = \alpha$

$x = \frac{ - 1}{4} = \beta$

Verification of relation between zeroes and co efficients :

$</u><u>\alpha + \beta = \frac{ - b}{a}$

$- 1 - \frac{1}{4} = - \frac{5}{4}$

$\frac{ - 4 - 1}{4} = \frac{ - 5}{4}$

$\frac{ - 5}{4} = \frac{ - 5}{4}$

LHS = RHS

$\alpha \beta = \frac{c}{a}$

$- 1 \times - \frac{1}{4} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

LHS = RHS

Hence, verified.

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