⟨1.⟩
Given equation is -4x-30 = -2x²
⇛ -4x-30+2x² = 0
⇛ 2x²-4x-30 = 0
⇛ 2(x²-2x-15) = 0
⇛ x²-2x-15 = 0
It is the form of ax²+bx+c = 0
We have,
a = 1
b = -2
c = -15
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
Discriminant = (-2)²-4(1)(-15)
⇛ D = 4+60
⇛ D = 64 > 0
The given equation has two distinct and real roots.
Now,
x²-2x-15 = 0
⇛ x²+3x-5x-15 = 0
⇛ x(x+3)-5(x+3) = 0
⇛ (x+3)(x-5) = 0
⇛ x+3 = 0 or x-5 = 0
⇛ x = -3 or x = 5
The roots are -3 and 5.
⟨2.⟩
Given equation is 3x²+27x = 0
It is the form of ax²+bx+c = 0
We have,
a = 3
b = 27
c = 0
The discriminant of ax²+bx+c = 0 is b²-4ac
Discriminant = 27²-4(3)(0)
⇛ D = 729 > 0
The given equation has two distinct and real roots.
⇛3x(x+9) = 0
⇛ 3x = 0 or x+9 = 0
⇛ x = 0/3 or x = -9
⇛ x = 0 or x = -9
The roots are 0 and -9.
⟨3.⟩
Given equation is x²+6x+9 = 0
It is the form of ax²+bx+c = 0
We have,
a = 1
b = 6
c = 9
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
Discriminant = 6²-4(1)(9)
⇛ D = 36-36
⇛ D = 0
The given equation has two equal and real roots.
Now,
x²+6x+9 = 0
⇛ x²+3x+3x+9= 0
⇛ x(x+3)+3(x+3) = 0
⇛ (x+3)(x+3) = 0
⇛ x+3 = 0 or x+3 = 0
⇛ x = -3 or x = -3
The roots are -3 and -3.
⟨4.⟩
Given equation is 4x²+16 = 8x
⇛ 4x²+16-8x = 0
⇛ 4x²-8x+16 = 0
⇛ 4(x²-2x+4) = 0
⇛ x²-2x+4 = 0
It is the form of ax²+bx+c = 0
We have,
a = 1
b = -2
c = 4
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
Discriminant = (-2)²-4(1)(4)
⇛ D = 4-16
⇛ D = -12 < 0
The given equation has no real roots.
⟨5.⟩
Given equation is x² +5x = 24
⇛ x²+5x-24 = 0
It is the form of ax²+bx+c = 0
We have,
a = 1
b = 5
c = -24
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
Discriminant = (5)²-4(1)(-24)
⇛ D = 25+96
⇛ D = 124 > 0
The given equation has two distinct and real roots.
Now,
x²+5x-24 = 0
⇛ x²+8x-3x-24 = 0
⇛ x(x+8)-3(x+8) = 0
⇛ (x+8)(x-3) = 0
⇛ x+8 = 0 or x-3 = 0
⇛ x = -8 or x = 3
The roots are -8 and 3.
Answer:
