Question

Prove that

${\boxed{\pink{\pmb{\mathfrak{ax^2+bx+c}}}}}$

Answer 1

ax2 + bx + c = 0

ax2 + bx = –c

Divide through by whatever is multiplied on the squared term.

x2 + (b / a) x  = - c / a

and simplify

x2 + (b / a) x = - c / a

Add (b / 2a)2 to both sides

x2 + (b / a) x + (b / 2a)2 = - c / a+ (b / 2a)2

to complete the square

(x + b / 2a )2 = - c / a + (b / 2a)2

Group the two terms on the right side of the equation

(x + b / 2a)2 = (b2 - 4 a c) / (4 a2 )

Solve by taking the square root

x + b / 2a = ± √{ (b2 - 4 a c ) / (4 a2) }

Solve for x to obtain two solutions

x = - b / 2a ± √{ (b2 - 4 a c ) / (4 a2) }

The term √{ (b2 - 4 a c ) / (4 a2) }may be simplified as follows

√{ (b2 - 4 a c ) / (4 a2) } = √(b2 - 4 a c) / (2 |a|)

Since 2 | a | = 2 a for a > 0 and 2 | a | = -2 a for a < 0, the two solutions to the quadratic equation may be written

x = (-b + √( b2 - 4 a c)) / (2 a)

x = (-b - √ ( b2 - 4 a c)) / (2 a)

Answer 2

Answer:

The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) .

Step-by-step explanation:

$a {x}^{2} + bx + c = 0$

We now identify the coefficients a, b and c.

a = 2 , b = 3 and c = 9

The sign of the discriminant Δ =

${b}^{2} - 4ac$

the number of solutions to the equation.

Δ =

${b}^{2} - 4ac = {3}^{2} - 4(2)(9) = - 63$

The discriminant Δ is negative and therefore the equation has two complex solutions.