Question

find the zero of the quadratic polynomial 2x*x+5x-12 and verify relationship between zeroes and the coefficient​

Answer 1

$\bigstar\large\underline{\pmb{\sf Correct \:Question }}$

find the zeros of the quadratic polynomial 2x²+5x-12 and verify the relationship between zeroes and the coefficient​.

$\bigstar\large\underline{\pmb{\sf Solution }}$

$:\implies\sf p(x)=2x^2+5x-12=0$

$:\implies\sf p(x)=2x^2+(8-3)x-12=0$

$:\implies\sf p(x)=2x^2+8x-3x-12=0$

$:\implies\sf p(x)=2x(x+4)-3(x+4)=0$

$:\implies\sf (2x-3)(x+4)=0$

$:\implies\sf \orange{\alpha=\dfrac{3}{2} ;\beta=-4}$

$\bigstar\large\underline{\pmb{\sf Verifying\:the\:zeros }}$

following are the two formulas to be used here:-

$\longrightarrow\underline{\boxed{\bf \alpha+\beta=\frac{-b}{a} }}$

$\longrightarrow\underline{\boxed{\bf \alpha\beta=\frac{c}{a} }}$

here, a, b and c are coefficient of x², coefficient of x, and constant respectively. In the given equation they are 2, 5, and -12 respectively.

$\dag\underline{\rm{\sf Sum\:of\:Zeros }}$

$:\implies\sf \alpha+\beta=\dfrac{-b}{a}$

$:\implies\sf \dfrac{3}{2}+(-4)=\dfrac{3-8}{2}=\red{\dfrac{-5}{2}}$

$:\implies\sf\dfrac{-b}{a}=\red{\dfrac{-5}{2}}$

$\therefore \sf LHS=RHS$

$\dag\underline{\rm{\sf Product\:of\:Zeros }}$

$:\implies\sf \bigg(\dfrac{3}{\cancel{2}}\bigg)(\cancel{-4}^{-2})=\red{-6}$

$:\implies\sf \dfrac{c}{a}=\dfrac{\cancel{-12}}{\cancel{2}}=\red{-6}$

$\therefore \sf LHS=RHS$

$\sf\pink{Hence, Verified!!}$

Answer 2

★VERIFYING ZEROS

= A+B = -a/b

= AB = c/a

★SUM OF ZEROS

= A+B = - b/a

= 3/2 + (-4) = 3-8/2 = -5/2

= -B/A = 5/2

LHS = RHS

★PRODUCT OF ZEROS

= (3/2) (-²) = -6

= C/A =2/2 = - 6

hope it will be helpful to you