Question

can you please tell how we derived

formulas in
boldened letters

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Answer 1

Question:-

Derivation of Sum & Product of the zeroes of a quadratic equation ax² + bx + c.

i.e., we have to prove that

• α + β = - b/a

• αβ = c/a

Answer:-

We know:

If x is the one of the zeroes of a Quadratic polynomial ,

$\sf \implies \: x = \dfrac{ - \: b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

So,

$\sf \implies \: \alpha = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \: \: - - \: \: equation \: (1)$

And,

$\implies \sf \: \beta = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \: \: - - \: \: equation \: (2)$

Add equations (1) & (2).

$\implies \sf \: \alpha + \beta = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} + \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \implies \sf \: \: \alpha + \beta = \frac{ - b + \cancel{\sqrt{ {b}^{2} - 4ac } } - b - \cancel{ \sqrt{ {b}^{2} - 4ac}} }{2a} \\ \\ \implies \sf \: \: \alpha + \beta = \frac{ - \cancel{ 2}b}{ \cancel{2}a} \\ \\ \implies \sf \red{ \: \alpha + \beta = \frac{ - b}{a} }$

Now,

Multiply equation (1) & (2).

$\: \implies \sf \: \: \alpha \beta = \bigg( \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \bigg) \bigg( \dfrac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} \bigg) \\ \\ \implies \sf \: \alpha \beta = \frac{( - b + \sqrt{ {b}^{2} - 4ac })( - b - \sqrt{ {b}^{2} - 4ac } ) }{2a \times 2a}$

using (a + b)(a - b) = a² - b² we get,

$\: \implies \sf \: \: \alpha \beta = \frac{( - b) ^{2} - ( \sqrt{ {b}^{2} - 4ac} ) ^{2} }{4 {a}^{2} } \\ \\ \implies \sf \: \alpha \beta = \frac{ {b}^{2} - {b}^{2} + 4ac }{4 {a}^{2} } \\ \\ \implies \sf \: \alpha \beta = \frac{ \cancel{4a}c}{ \cancel{4} {a}^{ \cancel 2} } \\ \\ \implies \sf \: \red{ \alpha \beta = \frac{c}{a} }$

Hence, derived that ,

• α + β = - b/a

• αβ = c/a