Question

3x² - x - 4
any one please slove this...
find the zeroes of the quadratic polynomial​

Answer 1

Step-by-step explanation:

3x ^2 - x - 4

3x ^2 - (4-3) x - 4

3x^2 - 4x + 3x - 4

3x^2 + 3x - 4x - 4

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

(3x - 4 ) (x+1)

❥ Hope it helps...

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Answer 2

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

• $3 {x}^{2} - x - 4$

$\color{red}\huge{\underline{\underline{\bf SOLUTION\dag}}}$

$\tt Suppose,$

$p(x) = 3 {x}^{2} - x - 4$

$\tt Let,$

$p(x) = 0$

$\implies 3 {x}^{2} - x - 4 = 0$

$\implies 3 {x}^{2} - 4x + 3x - 4 = 0$

$\implies 3x(x - 4) 1(x - 4) = 0$

$\implies (3x + 1)(x - 4) = 0$

$\tt two \: cases :$

$\tt case(i)$

$\implies 3x + 1 = 0$

$\implies 3x = - 1$

$\orange{ \therefore x = \frac{ - 1}{3} }$

$\tt case(ii)$

$\implies x - 4 = 0$

$\orange{ \therefore x = 4}$

Therefore zeros of quadratic polynomial are:

$x = 4 \: \: \: and \: \: \frac{ - 1}{3}$

HOPE THIS HELPS!!