If one root of the quadic polynomial ax²-6x-6 is 4 , then find the va

1.	If one root of the quadic polynomial ax²-6x-6 is 4 , then find the value of a . Also find the sum of its zeros
Answer» $\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$ If one root of the quadic polynomial ax²-6x-6 is 4 , then find the value of a . Also find the sum of its zeros ? $\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$ $\Large{\underline{\mathfrak{\bf{\red{Given}}}}}$ polynomial , ax² - 6x - 6 = 0 And, 4 is a zeroes of this equation $\Large{\underline{\mathfrak{\bf{\red{Find}}}}}$ Value of a Second zeroes of this equation . Sum of zeroes $\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$ We know, If 4 is a zeroes of this equation . so we can say that, this value of x exits of this equation. Keep value of x = 4 in this equation, ➠ a.(4)² - 6.(4) - 6 = 0 ➠ 16a - 24 - 6 = 0 ➠ 16a = 30 ➠ a = 30/16 ➠ a = 15/8 Now , keep value of a = 15/8 in this equation ➠ (15/8)x² - 6x - 6 = 0 ➠ 15x² - 48X - 48 = 0 ➠ 5x² - 16x - 16 = 0 (New equation) ➠ 5x² - 20x + 4x - 16 = 0 ➠ 5x (x - 4) + 4(x - 4 ) = 0 ➠ (5x + 4)(x-4) = 0 ➠ (5x + 4) = 0 Or, ( x -4) = 0 ➠ 5x = -4 Or, x = 4 ➠ x = -4/5 Or, x = 4 $\Large{\underline{\mathfrak{\bf{\red{Thus}}}}}$ Both zeroes of this equation is -4/5 and 4 Now, ➠ Sum of both zeroes = ( -4/5) + (4) ➠ Sum of both zeroes = (-4+20)/5 ➠ Sum of both zeroes = 16/5 $\Large{\underline{\mathfrak{\bf{\orange{Verification}}}}}$ We know, $\small\boxed{\sf{\green{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}}$ $:\mapsto\sf{\:Sum\:of\:zeroes\:=\:\dfrac{-(-16)}{5}} \\ \\ :\mapsto\sf{\pink{\:Sum\:of\:zeroes\:=\:\dfrac{16}{5}}}$ That's proved