Question

find the zeroes of polynomial 2x^2+5x-12 and verify the relation between zeroes ans cofficients​

Answer 1

Answer:

Step-by-step explanation:

2x^2 +5x -12

Split the middle term

2x^2+8x  -3x -12

2x[x+4] - 3[x+4]

[2x-3] [x+4]

zeroes are

= x= 3/2 or x= -4

verification

Let

3/2 be α and -4 be β

equation is written in the form ax^2+bx+c

α+β=-b/a

αβ= c/a

3/2 +(-4) = -5/2

3/2-8/2

=-5/2 = -5/2....(1)

3/2x-4= -12/2

-6= -6

HENCE VERIFIED

Answer 2

Solution :

We have quadratic polynomial p(x) = 2x² + 5x - 12

Zero of the polynomial p(x) = 0

A/q

$\longrightarrow\tt{2x^{2} +5x-12=0}\\\\\longrightarrow\tt{2x^{2} +8x-3x-12=0}\\\\\longrightarrow\tt{2x(x+4)-3(x+4)=0}\\\\\longrightarrow\tt{(x+4)(2x-3)=0}\\\\\longrightarrow\tt{x+4=0\:\:\:Or\:\:\:2x-3=0}\\\\\longrightarrow\tt{x=-4\:\:\:Or\:\:\:2x=3}\\\\\longrightarrow\bf{x=-4\:\:Or\:\:x=3/2}$

∴ We have α = -4 & β = 3/2 are the zeroes of the polynomial.

As we know that given polynomial compared with ax² + bx + c;

• a = 2
• b = 5
• c = -12

Now;

$\underline{\large{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x }{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\sf{-4+\dfrac{3}{2} =\dfrac{-5}{2} }\\\\\\\mapsto\sf{\dfrac{-8+3}{2} =\dfrac{-5}{2}} \\\\\\\mapsto\bf{\dfrac{-5}{2} =\dfrac{-5}{2} }$

$\underline{\large{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\sf{-4\times \dfrac{3}{2} =\dfrac{-12}{2} }\\\\\\\mapsto\sf{\cancel{\dfrac{-12}{2}} =\cancel{\dfrac{-12}{2}}} \\\\\\\mapsto\bf{-6=-6}$

Thus;

The zeroes of the polynomial & coefficient is verified .