Please help-I'm stuck really bad
What are the roots of the equation x^2 - 4x + 3 = 0
I think I was suppose to use b^2 - 4ac
so it would be -4 - (-4)*1*3 = 12 so the roots would be 12 and -12
Thank you for any help you can provide
x^2 -4x + 3 = 0
x^2 - 4x = 3
x^2 +2(-2)x + -2^2 = -3 = -2^2
x^2 + -4x + 4 = 1
(x + -2)^2 = 1
qrt (x+-2)^2 = sqrt 1
x+-2 = plus or minus 1
therefore x=1
x=3 roots are 1 and 3
I'm here to help you! Let's find the roots of the equation x^2 - 4x + 3 = 0 step by step.
To find the roots of a quadratic equation of the form ax^2 + bx + c = 0, you can use the quadratic formula or factorization. In this case, we will use factorization.
The given equation is x^2 - 4x + 3 = 0. To factorize it, we need to find two numbers that multiply to give the constant term (3) and add up to give the coefficient of the middle term (-4).
In this case, the numbers are -1 and -3 because (-1) * (-3) = 3, and (-1) + (-3) = -4.
Now, rewrite the equation by splitting the middle term using these two numbers:
x^2 - 1x - 3x + 3 = 0
Next, group the terms:
(x^2 - 1x) + (-3x + 3) = 0
Factor out common terms from each group:
x(x - 1) - 3(x - 1) = 0
Now, notice that we have a common factor of (x - 1). Factor this out:
(x - 1)(x - 3) = 0
The equation is now factored into two binomial expressions.
To find the roots, set each factor equal to zero:
x - 1 = 0 or x - 3 = 0
Solve these two equations separately:
x = 1 or x = 3
The roots of the equation x^2 - 4x + 3 = 0 are x = 1 and x = 3.
Therefore, the roots are not 12 and -12 as you mentioned. Be careful with the calculation step. It seems you made a mistake when applying the quadratic formula. Remember that the discriminant (b^2 - 4ac) should not be used to find the roots directly. Instead, it helps determine the nature of the roots (real, imaginary, etc.). In this case, the equation has two real and distinct roots.
I hope this explanation helps! Let me know if you have any further questions.