Question

find the zeros of 64x²+16x+1 and verify the relationship between the zeros and it's cofficient​

Answer 1

Answer:

$64x {}^{2} + 16x + 1 = 0 \\ 64x {}^{2} + 8x + 8x + 1 = 0 \\ 8x(8x + 1) + 1(8x + 1) = 0 \\ (8x + 1)(8x + 1) = 0 \\ \\ a = 64 \: \: \: \: b = 16 \: \: \: \: \: c = 1 \\ \\ x = - \frac{1}{8} = \alpha \\ \\ x = - \frac{1}{8} = \beta \\ \\ sum \: of \: zeroes \: \\ \alpha + \beta = - \frac{b}{a} \\ \\ - \frac{1}{8} - \frac{1}{8} = - \frac{16}{64} \\ \\ - \frac{1}{4} = - \frac{1}{4} \\ \\ product \: of \: zeroes \\ \alpha \beta = \frac{c}{a} \\ \\ - \frac{1}{8} \times - \frac{1}{8} = \frac{1}{64} \\ \\ \frac{1}{64} = \frac{1}{64}$

Step-by-step explanation:

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Answer 2

Answer:

⋆ Given Polynomial : 64x² + 16x + 1

Here : a = 64,⠀b = 16,⠀c = 1

$:\implies\tt f(x) = 0\\\\\\:\implies\tt 64x^2 +16x +1= 0\\\\\\:\implies\tt (8x)^2 +(2 \times 8x \times 1) +(1)^2= 0\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{(a)^2+2ab+(b)^2=(a+b)^2}}\\\\:\implies\tt (8x+1)^2 = 0\\\\\\:\implies\tt 8x+1=0\\\\\\:\implies\underline{\boxed{\tt x =\dfrac{-\:1}{8}}}$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{ -1}{8} + \dfrac{ -1}{8} = \dfrac{ -16}{64}\\\\\\\dashrightarrow\tt\:\: \dfrac{ - 2}{8} = \dfrac{ - 1}{4} \\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt \dfrac{ - \:1}{4} = \dfrac{ -\:1}{4} }}}\\\\\\{\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{ - 1}{8} \times \dfrac{ - 1}{8} = \dfrac{1}{64}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt \dfrac{1}{64} = \dfrac{1}{64} }}}$