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To solve using completing square method we proceed as follows:
x^2-10x+8=0
x^2-10x=-8
but
c=(b/2)^2
c=(10/2)^2=25
thus we can add this in our expression to get
x^2-10x+25=8+25
factorizing the LHS we get:
(x-5)(x-5)=33
(x-5)^2=33
getting the square roots of both sides we have:
x-5=+/-√33
x=5+/-√33
Equation :
[x(x+4)]/2=6 (as (b*h)/2=A)
x*x + 4*x -12 = 0
so, x = 2 units(as -6 is negative)
Answer:
566
Step-by-step explanation:
566
Answer:
Step-by-step explanation:
2(6x² - 3) = 11x² - x
2*6x² - 2*3 = 11x² - x
12x² - 6 = 11x² -x
Subtract 11x² from both sides
12x² - 11x² - 6 = -x
x² - 6 = -x
x² + x - 6= 0
Sum =1
Product = -6
Factors = 3 , (-2) { 3*(-2) = -6 & 3 +(-2) = 1}
x² + 3x - 2x - 6 = 0
x(x + 3) - 2(x + 3)= 0
(x +3)(x - 2) = 0
