Question

if 3 is a solution of 3 X square + the bracket K - 1 (X + 9 equal to zero find the value of k​

Answer 1

$3 {x}^{2} + (k - 1)x + 9 = 0$

so now put the value of x as 3

then,

$3 ({3})^{2} + (k - 1)3 + 9 = 0 \\ = (3 \times 9) + 9 + (k - 1)3 = 0 \\ = 27 + 9 +(k - 1)3 = 0 \\ = 36 + (k - 1)3 = 0 \\ = k - 1 = \frac{ - 36}{3} \\ = k = - 12 + 1 \\ = k = - 11(ans).$

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Answer 2

Answer:

Step-by-step explanation

3x square + k-1(x+9) =0

using b square minus 4ac which is called the discriminant of the quadratic equation

expand your equation which is 3x square + kx -9 =0

a=3; b= k;c= -9

using b square minus 4ac =0

(k) square -(4*3*-9) =0

k square +108 =0

when you take 108 to the right hand side it become minus

k square =-108

square root of k square = + or minus square root of 108

the square from the square cancels the square on top of k

square root of 108 becomes 6 root 3 which was gotten from root 36 and root 3 cause 36*3 gives 108

square root of 36 is 6

k= + or - 6 root 3.