The List II gives polynomial (quadratic) for the zeroes given in the List I, match them correctly.

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Solution

$x^{2} - 25 = 0$
$x^{2} - 5^{2} = 0$
$(x + 5) (x - 5) = 0$
$x = - 5$ and $x = 5$
Hence, D $\to$ 1.
$x^{2} + 2 x - 15 = 0$
$(x - 3) (x + 5) = 0$
$x = 3$ or $x = - 5$
Hence, A $\to$ 2.
$x^{2} + (\frac{80}{9}) x - 1 = 0$
$(x + 9) (x - \frac{1}{9}) = 0$
$x = - 9$ and $x = \frac{1}{9}$
Hence, C $\to$ 3.
$x^{2} - 10 x + 21 = 0$
$(x - 3) (x - 7) = 0$
$x = 3$ and $x = 7$
Hence, B $\to$ 4.

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The List II gives polynomial (quadratic) for the zeroes given in the List I, match them correctly.

x2−25=0x2−52=0(x+5)(x−5)=0x=−5 and x=5Hence, D →1.x2+2x−15=0(x−3)(x+5)=0x=3 or x=−5Hence, A→ 2.x2+(809)x−1=0(x+9)(x−19)=0x=−9 and x=19Hence, C→ 3.x2−10x+21=0(x−3)(x−7)=0x=3 and x=7Hence, B→4.