Find the roots of the quadratic equation if they exist by Completely the square x² -5x + 10 = 0
Answers
x² - 5x + 10 = 0
=> x² - 5x = -10
=> x² - 2.(5/2).x = -10
=> x² - 2.(5/2).x + (5/2)² = -10 + (5/2)²
=> (x - 5/2)² = -10 + 25/4 = (-40 + 25)/4
=> (x - 5/2)² = -15/4
here we see LHS is positive term {square of anything be always positive} while RHS is negative . hence, there is no real root of given quadratic equation.
METHOD OF COMPLETING THE SQUARE :
Step 1 - Write the given equation in standard form, ax²+bx+c = 0, a≠0.
Step 2 - If the coefficient of x² is 1, go to step 3. If not, divide both sides of the equation by the coefficient of x²
Step 3 - Shift the constant term (c/a) on RHS.
Step 4- Find half the coefficient of x and square it. Add this number to both sides of the equation.
Step 5 - Write LHS in the form a perfect square and simplify the RHS.
Step 6 - Take the square root on both sides.
step 7 : Find the values of x by shifting the constant term(b/2a) on RHS from LHS.
SOLUTION :
Given :
x² - 5x + 10 = 0
=> x² - 5x = -10
=> x² - 5x + (5/2)² = -10 + (5/2)²
=> x² - 2×(5/2)× x + (5/2)² = -10 + (5/2)²
=> (x - 5/2)² = -10 + 25/4
[(a² -2ab + b² = (a -b)²]
=> (x - 5/2)² = (- 40 + 25)/4
=> (x - 5/2)² = -15/4
which is not possible at the square of a real number cannot be negative.
Hence, roots of the equation x² - 5x + 10 = 0 does not exist .
HOPE THIS ANSWER WILL HELP YOU...