Question

x^(2)-7x+10





find the zeros of the following Quadratic polynomial and verify the relationship between zeros and coefficient ​

Answer 1

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Answer 2

$\bf ➯p(x) = {x}^{2} + 7x + 10$

$\bf ➣let$

$\bf ⇒{x}^{2} + 7x + 10 = 0$

$\bf ⇒{x}^{2} + 2x + e+ 10 = 0$

$\bf ⇒x(x + 2) + 5(x + 2) = 0$

$\bf➾(x + 5)(x + 2) = 0$

$➢So$

$\bf ➾x = ( - 2) \: and \: ( - 5)$

$\bf ➣Therefore$

$\bf ⇒a = - 2 \: and \: b = - 5$

$\bf ➯sum \: of \: zeros \: = \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\bf ⇒a + b = \dfrac{ - b}{a}$

$\bf ⇒- 2 + ( - 5) = \dfrac{ - 7}{1}$

$\bf ⇒- 7 = - 7$

$\bf ➯product \: of \: zeroes = \dfrac{c}{d}$

$\bf ⇒a \times b = \dfrac{c}{d}$

$\bf⇒( - 2)( - 5) = \dfrac{10}{1}$

$\bf ➾10 = 10$

Since,

L.H.S = R.H.S

Hope it helps you...