One of the solutions to the equation 5x^2+bx+12=0, is -4/5. find the other solution.

Question

hi! algebra help :) thorough explanation please if possible

Answer 1

Yeah, the answer by other tutor is one of the methods.

Below is another method to get b.

When roots are given to get the quadratic equation :

For a quadratic equation ax²+bx+c = 0, the sum of its roots = –b/a and the product of its roots = c/a

let the roots be 'm' and 'n'

Given one root m = - 4/5

5x^2 + bx + 12 = 0

the sum of its roots = –b/a

m + n = - b/a

-4/5 + n = -b/5

-4 + 5n = -b

5n= -b + 4 -------------1

product of its roots = c/a

m*n = c/a

-4/5 * n = 12/5

-4 n = 12

n = -3 is the other root -------------2

Plug-in equation 2 in 1

5(-3) = -b + 4

-15 = -b + 4

b = 4+15

b =19

Hope this helps! For any further doubts you can message me, i am here to clarify

Answer 2

5x^2+bx+12=0

substitute -4/5 for each x to find the value of "b"

5(-4/5)^2+b(-4/5)+12=0

5(16/25)-(4/5)b+12=0

(16/5)-(4b/5)+12=0

add 12 and 16/5 (3 1/5) to get 15 1/5

15 1/5 - (4b/5)=0

76/5=4b/5

because the denominators are the same the numerators must be the same

4b=76

b=19

substitute

5x^2+19x+12=0

factor

5*12=5*2*2*3=(5*3)*(2*2)=15*4 and 15+4=19, the middle term's coefficient

(5x+4)(x+3)=0

5x+4=0

5x=-4

x=-4/5

x+3=0

x=-3 is the other solution