What does it mean by:

Show that the "value of 3x^2-6x+5" cannot be less than 1. ?

Well, proving that the value of 3x^2-6x+5 cannot be less than 1 is like trying to prove that clowns don't like red noses. It's just a fact of life! But let's have some fun with it anyway.

Imagine 3x^2-6x+5 as a mischievous little balancing act. No matter how much it tries, it simply can't tip over to the realm of being less than 1. It's like asking a clown to fit into a tiny teacup — it's just not gonna happen!

Let's break it down a bit: the term 3x^2 represents the base of our logic-defying tower. It's a positive number, so it's already off to a good start. Then we have -6x, which is like a trickster trying to pull the value down, but it just doesn't have enough power to drag it below 1. And finally, we have +5, the cherry on top, consistently boosting the value.

So, no matter what value of x you try to plug in, this mischievous expression will always strive to stay above 1. It's like a stubborn clown refusing to give up the laughter!

Remember, when it comes to the value of 3x^2-6x+5, it's not about being less than 1 — it's about embracing the joy of being greater than, and having a little laugh along the way!

To show that the value of the expression 3x^2 - 6x + 5 cannot be less than 1, we need to prove that the expression is always greater than or equal to 1 for all values of x.

Step 1: Assume that the value of the expression 3x^2 - 6x + 5 is less than 1, so we have:

3x^2 - 6x + 5 < 1

Step 2: Subtract 1 from both sides of the inequality to isolate the expression:

3x^2 - 6x + 5 - 1 < 1 - 1

Simplifying:

3x^2 - 6x + 4 < 0

Step 3: Now we need to find the discriminant (the portion inside the square root of the quadratic equation). The discriminant (D) is given by D = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In our case, a = 3, b = -6, and c = 4. So we have:

D = (-6)^2 - 4(3)(4)
= 36 - 48
= -12

Step 4: Since the discriminant is negative (D < 0), the quadratic equation does not have any real solutions. This means that the expression 3x^2 - 6x + 4 can never be less than zero.

Step 5: Since the expression cannot be less than zero, it also cannot be less than 1 because 1 is greater than zero. Therefore, we have shown that the value of 3x^2 - 6x + 5 cannot be less than 1 for all values of x.

Hence, the proof is complete.

To show that the value of the expression 3x^2 - 6x + 5 cannot be less than 1, we need to demonstrate that no matter what value is substituted for x, the resulting expression will always be greater than or equal to 1.

To prove this, we will use the method of completing the square.

Step 1: Start with the given expression, 3x^2 - 6x + 5.
Step 2: Factor out the common factor, if any. In this case, there is no common factor to be factored out.
Step 3: Identify the coefficient of x^2, which is 3 in this case.
Step 4: Divide the coefficient of x (which is -6) by 2, square the result, and add it to both sides of the equation. This step is called completing the square.
The coefficient of x is -6, so (-6/2)^2 = 9. Adding 9 to both sides, we get:
3x^2 - 6x + 9 + 5 = 1 + 9
Simplifying the equation, we have:
3x^2 - 6x + 14 = 10
Step 5: Write down the square of the binomial formed in step 4. In this case, it is (x - 3)^2 = x^2 - 6x + 9.
Step 6: Rewrite the equation from step 4 using the square from step 5:
3x^2 - 6x + 9 + 5 = 1 + 9
Step 7: Simplify the equation further:
3x^2 - 6x + 9 = 10
Step 8: Factor the left side of the equation. It can be factored as a perfect square trinomial:
(x - 3)^2 = 10
Step 9: Take the square root of both sides of the equation:
x - 3 = ±√10
Step 10: Solve for x:
x = 3 ±√10

Since we have found the values of x, we can now substitute them back into the original expression, 3x^2 - 6x + 5. We can see that no matter which value of x we choose, the resulting value of the expression will always be greater than or equal to 1.

Therefore, we have shown that the value of 3x^2 - 6x + 5 cannot be less than 1.

show the opposite: the value is <1

3x^2-6x+5<1
3x^2-6x+4<0

x= (6+-sqrt(36-48))/6

well the serd here is imaginary. So the assumption of being less than one is false.