Question

Please answer this question.. correct answer will get brainliest

Answer 1

Step-by-step explanation:

Solution in Attachment.

Answer 2

Step-by-step explanation:

$Let \: \alpha = \frac{1}{2} \\: \beta = \frac{-3}{2}\: and\:\\ p(x)= 4x^{2}+4x-3 --(1)$

Now ,

$i ) Put\: \alpha\: in \:equation\: (1), \:we \:get\\p(\frac{1}{2})=4(\frac{1}{2})^{2}+4(\frac{1}{2})-3\\=4\times \frac{1}{4}+2-3\\=1+2-3\\=0$

Therefore,

$p(\frac{1}{2})=0$

$\alpha \: is \: a zero\: of \:p(x)$

$ii) Put\: \beta\: in \:equation\: (1), \:we \:get\\p(\frac{-3}{2})=4(\frac{-3}{2})^{2}+4(\frac{-3}{2})-3\\=4\times \frac{9}{4}-6-3\\=9-6-3\\=0$

Therefore,

$p(\frac{-3}{2})=0$

$\beta \: is \: a zero\: of \:p(x)$

iii) Compare 4x²+4x-3 with

ax²+bx+c , we get

a=4, b=4, c =-3

$Sum \:of \:the \: zeroes \\=\frac{1}{2}+\frac{-3}{2}\\=\frac{1-3}{2}\\=\frac{-2}{2}\\=\frac{-1}{1}\\=\frac{-coefficient\:of\:x}{coefficient\:of \:x^{2}}$

$Product\:of\:zeroes\\=\frac{1}{2}\times \frac{-3}{2}\\=\frac{-3}{4}=\frac{constant}{coefficient\:of\:x^{2}}$

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