Hello:) I need help with 11(ii). Question in the pic above:) Thank you in advance.

it has 3 real roots!

Step-by-step explanation:

Factor of the equation is (x+1) &(x^2-4x-2)

since x+1 is it's factor so it's one root is x+1=0

i.e,x= -1

for other roots we have to factorise the quadrrstic equation

x^2-4x+2 , it's Discriminant is (-4)^2-4×2=8, it's positive , that means it have real roots

it's root is x= (4±√8)/2. [using roots equal to -b±√b^2-4ac/2a]

i.e, root is 2±√2 {they are real numbers}

i.e, quad eqn is {x-(2+√2)}{x-(2-√2)},

✌️:)

Q11. i) f(x)=

ii) 3 real roots

Step-by-step explanation:

Please see the attached pictures for full solution.

For part (i), you forgot to multiply 2 with -2x. To check your answer, substitute -1 into f(x) since it is stated that -1 is a root of the equation f(x)=0. Ensure that the result is 0.

(ii) Factorise f(x) with the help of part (i) and the question. Note that it is ok to divide by 2 throughout for the third line of the working since 2 is a constant.

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.

step-by-step explanation: