Explanation:
⟨1.⟩
Given equation is 2x²+9x = -4
⇛ 2x²+9x+4 = 0
The standard form of the equation is 2x²+9x+4 = 0
On comparing the given equation with ax²+bx+c = 0
We have
a = 2
a = 9
c = 4
By using Quadratic Formula
x = [-b±√(b²-4ac)]/2a
On Substituting these values in the above formula then
⇛ x = [-9±√{9²-4(2)(4)}]/2(2)
⇛ x = {-9±√(81-32)}/4
⇛ x = {-9±√49}/4
⇛ x = (-9±7)/4
⇛ x = (-9+7)/4 or (-9-7)/4
⇛ x = -2/4 or -16/4
⇛ x = -1/2 or -4
The roots are -1/2 and -4
⟨2.⟩
Given equation is 3(x-4)²+11 = 0
⇛ 3(x²-8x+16)+11 = 0
⇛ 3x²-24x+48+11 = 0
⇛ 3x²-24x+59 = 0
The standard form of the equation is
3x²-24x+59= 0
On comparing the given equation with ax²+bx+c = 0
We have
a = 3
= 3b = -24
= 3b = -24c = 59
By using Quadratic Formula
x = [-b±√(b²-4ac)]/2a
On Substituting these values in the above formula then
⇛ x = [-(-24)±√{(-24)²-4(3)(59)}]/2(3)
⇛ x = {24±√(576-708)}/6
⇛ x = {24±√-132}/6
⇛ x = {24±2(-33)}/6
⇛ x = 2(12±√-33)/6
⇛ x = (12±√-33)/3
The roots are (12+√-33)/3 and
(12-√-33)/3
⟨a.⟩
I used the Quadratic Formula bu substituting the values of a , b and c values in it.
⟨b.⟩
Given two equations are Quadratic equations so they have two solutions or roots.
Since the degree of the Quadratic equations is 2.
Hope this helps!!
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