Question

9. Find the zeroes of the quadratic polynomial
$x {}^{2} + 43x + 222.$

Standard:- 10

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Answer 1

Hey!!!!

Good Evening

Difficulty Level : Average

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We have

=> x² + 43x + 222 = 0

Here 222 = 6 x 37

Thus by middle term splitting method

=> x² + 37x + 6x + 222 = 0

=> x(x + 37) + 6(x + 37) = 0

=> (x + 37)(x + 6) = 0

Thus x = - 37 or x = - 6 <<<<<<<<<< Answer

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Hope this helps ✌️

Answer 2

$\large {\mathbf {\red {Hello!!}}}$

$\blue {\bold {Here \: is \:the \: answer}}$

Given :
$x {}^{2} + 43x + 222$

By breaking the mid term....

We have to find two numbers such that their sum = 43 and product = 222.

By looking at the$\blue{Prime \: factorisation}$ of 222,
222 = 2×3×37.
we get two numbers as 6 and 37.

So,
⏩ x^2 + 43x + 222 = 0
⏩ x^2 + 37x + 6x + 222 = 0
⏩ x (x + 37) + 6 ( x + 37) = 0
⏩ (x + 6)(x + 37) = 0

By comparing with zero...
x + 6 = 0 and x+37 = 0
x = -6 and x = -37.

$\large {\mathbf {\red {HOPE \: \: IT \:\: HELPS !!}}}$