Question

Find the zeros of the quadratic polynomial x2 +3x +2 and verify the relationship between the zeros and the co-efficient

Answer 1

$hey \: mate \: here \: is \: your \: answer...$

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${x}^{2} + 3x + 2$
$x {}^{2} + (2 + 1)x + 2$
${x}^{2} + 2x + x + 2$
$x(x + 2) + 1(x + 2)$
$(x + 1)(x + 2)$
$(x + 1) = 0 \: \: \: \: \: \: \: \: \: \: \: \: (x + 2) = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: x = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:x = - 2$



$verification - - - -$

$sum \: of \: zeros \: = \frac{b}{a} \\ \\ \: \: \: \: \: \: \: - 1 + ( - 2) = \frac{ - 3}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 1 - 2 = - 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 3 = - 3$


$product \: of \: zeros = \frac{c}{a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 1 \times - 2 = \frac{2}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:2 = 2$


$hence \: verified..$



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$hope \: it \: helps....$

Answer 2

The zeroes of the given quadratic polynomial are 1, 2.

• Given polynomial is    

                  $x^{2} + 3x + 2$            -----------------------(1)

• If $x_{1}, x_{2}$ are roots of a polynomial, then the polynomial is written as

                  $x^{2} - (x_{1} + x_{2} )x + x_{1} x_{2}$  -----------------------(2)

• Writing (1) in the form of (2), we get

                  $x^{2} - (-2 - 1)x + 2$

• Comparing a and b, we get

                  $x_{1} = -2, x_{2} = -1$

• Hence, the roots of given polynomial are 2, 1.
• Verification of relationships:
• (i) Sum of roots =  $\frac{- coefficient of x}{coefficient of x^{2}}$ =

             -2-1 = -3/1

               -3 = -3

• (ii) Product of roots = $\frac{constant}{- coefficient of x^{2}}$    

                (-2)(-1) = 2/1

                   2 = 2