How do I solve this using the Quadratic formula and finding the determinant?
1. 5x-6=0
2. x+25=0
Neither equation you state is a quadratic equation
for the first, we could attempt to write it as
0x^2 + 5x - 6 = 0 , but when we attempt to use the formula x = (-b ± √(b^2 - 4ac) )/2a
we get x = (-5 ± √(25 - 0)/0
= 0/0 , which of course causes all kinds of problems
did you check that you typed it correctly ?
Step 1:
5x-6=0
+6 +6 Add the six to both sides
Step 2:
Then divide the five to both sides isolate the x
5x=6
/5 /5
Answer should be:
x=6/5
Then do the same to the second one and you should get
x=-25 if you did it right
To solve these equations using the quadratic formula and finding the determinant, we need to rewrite them in the form ax^2 + bx + c = 0.
1. Let's solve the equation 5x - 6 = 0 using the quadratic formula and finding the determinant.
Step 1: Rewrite the equation in the form ax^2 + bx + c = 0.
In this case, we have 5x - 6 = 0. We need to isolate the variable on one side, so adding 6 to both sides, we get 5x = 6.
Step 2: Further isolate the variable by dividing both sides by the coefficient of x.
Dividing both sides by 5, we get x = 6/5.
2. Let's solve x + 25 = 0 using the quadratic formula and finding the determinant.
Step 1: Rewrite the equation in the form ax^2 + bx + c = 0.
In this case, we have x + 25 = 0. To isolate x, we subtract 25 from both sides, giving us x = -25.
In both cases, you can notice that the equations are linear equations, not quadratic equations. The quadratic formula and finding the determinant are specific techniques used to solve quadratic equations, where the highest power of x is 2. However, these techniques are not necessary for linear equations like the ones given in your question.