solve 5x^2-17x+16 find the value of x
Answers
we should equal the value to 0
so there should be 0.
Then we can find value of x.
please verify the question
5x2-17x+16=0
Two solutions were found :
x =(17-√-31)/10=(17-i√ 31 )/10= 1.7000-0.5568i
x =(17+√-31)/10=(17+i√ 31 )/10= 1.7000+0.5568i
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(5x2 - 17x) + 16 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 5x2-17x+16
The first term is, 5x2 its coefficient is 5 .
The middle term is, -17x its coefficient is -17 .
The last term, "the constant", is +16
Step-1 : Multiply the coefficient of the first term by the constant 5 • 16 = 80
Step-2 : Find two factors of 80 whose sum equals the coefficient of the middle term, which is -17 .
-80 + -1 = -81
-40 + -2 = -42
-20 + -4 = -24
-16 + -5 = -21
-10 + -8 = -18
-8 + -10 = -18
For tidiness, printing of 14 lines which failed to find two such factors, was suppressed
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 2 :
5x2 - 17x + 16 = 0
Step 3 :
Parabola, Finding the Vertex :
3.1 Find the Vertex of y = 5x2-17x+16
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 5 , is positive (greater than zero).
Function has no real roots
Solve Quadratic Equation by Completing The Square
3.2 Solving 5x2-17x+16 = 0 by Completing The Square .
Divide both sides of the equation by 5 to have 1 as the coefficient of the first term :
x2-(17/5)x+(16/5) = 0
Subtract 16/5 from both side of the equation :
x2-(17/5)x = -16/5
Now the clever bit: Take the coefficient of x , which is 17/5 , divide by two, giving 17/10 , and finally square it giving 289/100
Add 289/100 to both sides of the equation :
On the right hand side we have :
-16/5 + 289/100 The common denominator of the two fractions is 100 Adding (-320/100)+(289/100) gives -31/100
So adding to both sides we finally get :
x2-(17/5)x+(289/100) = -31/100
Adding 289/100 has completed the left hand side into a perfect square :
x2-(17/5)x+(289/100) =
(x-(17/10)) • (x-(17/10)) =
(x-(17/10))2
Things which are equal to the same thing are also equal to one another. Since
x2-(17/5)x+(289/100) = -31/100 and
x2-(17/5)x+(289/100) = (x-(17/10))2
then, according to the law of transitivity,
(x-(17/10))2 = -31/100
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(17/10))2 is
(x-(17/10))2/2 =
(x-(17/10))1 =
x-(17/10)
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
x-(17/10) = √ -31/100
Add 17/10 to both sides to obtain:
x = 17/10 + √ -31/100
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Since a square root has two values, one positive and the other negative
x2 - (17/5)x + (16/5) = 0
has two solutions:
x = 17/10 + √ 31/100 • i
or
x = 17/10 - √ 31/100 • i
Note that √ 31/100 can be written as
√ 31 / √ 100 which is √ 31 / 10
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving 5x2-17x+16 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 5
B = -17
C = 16
Accordingly, B2 - 4AC =
289 - 320 =
-31
Applying the quadratic formula :
17 ± √ -31
x = ——————
10
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly,√ -31 =
√ 31 • (-1) =
√ 31 • √ -1 =
± √ 31 • i
√ 31 , rounded to 4 decimal digits, is 5.5678
So now we are looking at:
x = ( 17 ± 5.568 i ) / 10
Two imaginary solutions :
x =(17+√-31)/10=(17+i√ 31 )/10= 1.7000+0.5568i
or:
x =(17-√-31)/10=(17-i√ 31 )/10= 1.7000-0.5568i
Two solutions were found :
x =(17-√-31)/10=(17-i√ 31 )/10= 1.7000-0.5568i
x =(17+√-31)/10=(17+i√ 31 )/10= 1.7000+0.5568i