Question

Find the zeroes of the quadratic Polynomial 3x2 – x – 4 and verify the relationship between
the zeroes and the coefficient.

Answer 1

Answer:

4/3 ,  -1

Step-by-step explanation:

3x²-x-4 =0

3x²-4x+3x-4 =0

x(3x-4) + 1(3x-4) = 0

(3x-4) (x+1) = 0

x = 4/3  ,   x= -1

Answer 2

Given :-

$\: \mapsto \tt \: p(x) = 3 {x}^{2} - x - 4 \\ \\ \mapsto \tt \: 3 {x}^{2} - (4 - 3)x - 4 \\ \\ \tt \mapsto3 {x}^{2} - 4x + 3x - 4 \\ \\ \tt \mapsto \: x(3x - 4) + 1(3x - 4) \\ \\ \tt \mapsto \: (3x - 4)(x + 1)$

Now ,

$\tt \: \mapsto \: 3x - 4 = 6 \\ \\ \tt \: \mapsto3x = 4 \\ \\ \tt \: \mapsto \: x = \frac{4}{3}$

And ,

$\tt \: \mapsto \: x + 1 = 0 \\ \\ \tt \: \mapsto \: x = ( - 1)$

$\tt \: the \: two \: zeros \: of \: p(x) \: are \: \frac{4}{3} \: and \: - 1$

We know that ,

$\tt \: \dashrightarrow \: \frac{4}{3} = 1$

So ,

$\tt \: \rightarrow \: \: \frac{ - coefficient \: of \: x \: }{coeffficient \: of \: {x}^{2} } \\ \\ \tt \: \rightarrow \: \frac{1}{3} = \: sum \: of \: zeros \: \\ \\ \tt \: \rightarrow \: \: product \: of \: zeros \: ( \gamma \: \beta ) \\ \\ \tt \: \rightarrow \: \: \frac{4}{3} \times - 1 \: = \frac{ - 4}{3}$

So ,

$\tt \: \longmapsto \: \: \frac{constant}{coefficient \: of \: {x}^{2} } \\ \\ \tt \: \longmapsto \: \frac{ - 4}{3} = product \: of \: zeros$

Hence ,

• x = -1

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Hope it is helpful but if you have any doubt in my answer then let me know.

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