Question

the zeros of 5x^2-4-8x​

Answer 1

Step-by-step explanation:

We Have :-

5x²-8x-4

To Find :-

Zeroes of the polynomial

Method Used :-

Middle Term Splitting

Solution :-

$5x^{2} - 8x - 4 = 0\\ \\ \\ 5x^{2} - 10x + 2x - 4 =0 \\ \\ \\ 5x(x - 2) + 2(x - 2) = 0\\ \\ \\ (x - 2)(5x + 2) = 0 \\ \\ \\ so \: x \: can \: either \: be \: \\ \\ \\ x - 2 = 0 \\ \\ \\ x = 2 \\ \\ \\ or \\ \\ \\ 5x + 2 = 0 \\ \\ \\ 5x = - 2 \\ \\ \\ x = \dfrac{ - 2}{5}$

Answer 2

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❏ Question:-

Find the Zeroes of the polynomial ,

$\sf{\ \ {5x^2-4-8x}}$

❏ Solution:-

$\implies \sf{\ \ {5x^2-4-8x=0}}$

$\implies \sf{\ \ {5x^2-8x-4=0}}$

$\implies \sf{\ \ {5x^2-(10-2)x-4=0}}$

$\implies \sf{\ \ {5x^2-10x+2x-4=0}}$

$\implies \sf{\ \ {5x(x-2)+2(x-2)=0}}$

$\implies \sf{\ \ {(x-2)(5x+2)=0}}$

So Either ,

$\longrightarrow (x-2)=0$

$\longrightarrow \boxed{x=2}$

Or ,

$\longrightarrow (5x+2)=0$

$\longrightarrow 5x=-2$

$\longrightarrow \boxed{x=\dfrac{-2}{5}}$

∴ The Zeroes are $\longrightarrow x=2 \:and\:\dfrac{-2}{5}$

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