<span>x^2 + 15x + 56.25 = 105.25
"Completing the square" is one of many different techniques for solving a quadratic equation. What you do is add a constant to both sides of the equation such that the lefthand side can be factored into the form a(x+d)^2. For instance, squaring (X+D) = X^2 + 2DX + D^2. Notice the 2DX term. That is the same term as the 15x term in the problem. So 2D = 15, D = 7.5. And D^2 = 7.5^2 = 56.25.
So we have
x^2 + 15x + 56.25 = 49 + 56.25
Which is
x^2 + 15x + 56.25 = 105.25
Which is the answer desired.
Now the rest of this is going beyond the answer. Namely, it's answering the question "Why does complementing the square help?"
Well, we know that the left hand side of the equation can now be written as
(x+7.5)^2 = 105.25
Now take the square root of each side
(x+7.5) = sqrt(105.25)
And let's use both the positive and negative square roots.
So
x+7.5 = 10.25914226
and
x+7.5 = -10.25914226
And let's find X.
x+7.5 = 10.25914226
x = 2.759142264
x+7.5 = -10.25914226
x = -17.75914226
So the roots for x^2 + 15x - 49 is 2.759142264, and -17.75914226</span>
Answer:
<h2>The x-interecepts are 5.6 and -1.4, approximately.</h2>
Step-by-step explanation:
The given equation is
Where 
, 
and 
, let's use the quadratic formula
Therefore, the x-interecepts are 5.6 and -1.4, approximately.
Answer:
Step-by-step explanation:
50 - x² = 0
50 = x²
x² = 50
x = √50
Answer:
The expression is equal to 8
Step-by-step explanation:
This is the question in proper
6t−20−32u
When u= 1/4 t= 6
We were given this, 6t−20−32u then we were given the value of u= 1/4 and that of t= 6.
All we need here is to substitute u= 1/4 t= 6 into 6t−20−32u then evaluate it as follows
6(6)-20-32(1/4)
6×6=36
32×(1/4)=8
Then merge it together and calculate it we have
=36-20-8=8
