Question

Topic: The Quadratic Formula

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Match each equation on the left with the number and type of its solutions on the right.
x + 3x – 4= 0
two complex (nonreal) solutions
x2 + 3x + 4 = 0
two real solutions
4x² + 1 = 4x
one real solution
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Answer 1

Explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

`x= (-b +- √(b^2-4ac) )/(2a)`

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² = x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

`x= (-3 +- √(3^2-4(1)(-4)) )/(2(1)) \\= (-3+-√(25) )/(2) \\ = (-3+-5)/(2) \\= -8/2 or 2/2\\= -4 or 1`

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

`x= (-3 +- √(3^2-4(1)(4)) )/(2(1))\\= (-3 +- √(9-16) )/(2)\\= (-3 +- √(-7) )/(2)\\`

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

`x=(-(-4) +- √((-4)^2-4(4)(1)) )/(2(4)) \\= (4+-√(16-16) )/(8) \\= (4+-0)/(8) \\= 1/2`

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution