Question

Determine the solution set of each equation

Answer 1

Answer:

weak in maths sorry I cannot give answer

Answer 2

A=x1 = -8,x2 =2, B=x1= -5,x2=5, C=y1 = -4, y2 =5, D=y 18 over 5

alternative form

y=3 3over 5, =3.6

And E=x1=0, x2 =2 over 3

Alternative form

x1=0, x2=0.666667.

question A

Solving steps:Solve the quadratic equation

x²+6x-16=0

write 6x as a difference

x2 +8x-2x-16=0

factor out x for the expression

xx(x+8)-2x-16=0

factor out -2 from the expression

xx(x+8)- 2(x+8)=0

factor out x+8 from the expression

(x+8)x(x- 2)=0

when the product of factors equals 0,at least one factor is 0

x+8=0

x - 2=0

solve the equation for x

x= -8

x - 2=0

solve the equation for x

x= -8

x =2

the equation has 2 solution

x1 = -8,X2 =2

answer is

x1 = -8,x2 =2

Or

Solving steps:Number of solution

x²+6x-16=0

Determine the number of solution using the

discriminant D =b²- 4ac

D =6²- 4x1x(-16)

simply the expression

D=100

Since D>0, the quadratic equation has 2 real solution

2 real solution

Answer is

2 real solution

Question B

Solving steps:Solve the quadratic equation

x²-25=0

Move the constant to the right-hand side and change its sign

x²=25

Take the square root of both sides of the equation and remember to use both positive and negative roots

x=+5

Write the solutions,one with a +sign and one with a -sign

x= -5

x=5

The equation has 2 solutions

x1= -5,x2=5

the answer is

x1= -5,x2=5

Or

Solving steps:Number of solution

x²-25=0

Determine the number of solutions using the discriminant D=b²- 4ac

D=0²- 4x1x(-25)

Simply the expression

D=100

Since D>0,the quadratic equation has 2real solutions

Answer is

2real solutions

answer of question C is

y1 = -4, y2 =5

Answer of question D is

y 18 over 5

alternative form

y=3 3over 5, =3.6

Answer of question E is

x1=0, x2 =2 over 3

Alternative form

x1=0, x2=0.666667

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