Question 206219: 2x^2+20=-13x
Answer by anantha(86)   (Show Source):
You can put this solution on YOUR website! sol:
solving a Quadratic equation
step1: write the given Quadratic equation in standard form,ax^2+bx+c=0
step2: factorize the trinomial ax^2+bx+c and express it as the product of two linear factors
step3: put each linear factor equal to zero
thus,two values of x will be obtained,these are the roots of the given quadratic equation.
sol:
we have 2x^2+20= -13x
step1:given equation in standard form is 2x^2+20+13x=0
here -13x is transpose to L.H.S it convert into +13x
step2:
now we find the factors of this quadratic equation
2x^2+13x+20=0
now find two numbers with sum=13 and product=(2*20)=40
clearly,such numbers are 8 and 5,adding these numbers we get 8+5=13 and product we get 8*5=40
now 2x^2+13x+20=0 can be written as
2x^2+8x+5x+20=0
(2x^2+8x)+(5x+20)=0
2x(x+4)+5(x+4)=0
(x+4)(2x+5)=0
step3:
x+4=0 or 2x+5=0
x+4=0
x=-4 ,transpose +4 from L.H.S to R.H.S we get -4
2x+5=0
2x= -5
dividing by 2 on both sides
2x/2=-5/2
x= -5/2
x= -4 or -5/2
these are the roots of the quadratic equation
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