Question

find the zeros of P (x) = x² - 6x + 8 and verify the relationship between the zeros and cofficients
$(x) = x² - 6x + 8$
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Answer 1

${x}^{2} - 6x + 8 = 0$

Or,

${x}^{2} - 4x - 2x + 8 = 0$

Or,

$x(x - 4) - 2(x - 4) = 0$

Or,

$(x - 2)(x - 4) = 0$

Therefore,

Either x-2=0 or x-4=0.

$x - 2 = 0$

Or,

$x = 2$

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$x - 4 = 0$

Or,

$x = 4$

Therefore,

$x = 4$

$x = 2$

Check:-

$\alpha + \beta = \frac{ - b}{a}$

Now,

$\alpha + \beta = 6$

$\frac{ - b}{a} = \frac{ - ( - 6)}{1}$

$= 6$

Therefore,

$\alpha + \beta = \frac{ - b}{a}$

Now,

$\alpha \beta = \frac{c}{a}$

Here,

$\alpha \beta = 4 \times 2 = 8$

$\frac{c}{a} = \frac{8}{1}$

$= 8$

Therefore,

$\alpha \beta = \frac{c}{a}$

Hence, the relationship holds true.

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