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Quý An say: ....
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Niteesh say: First thing is, you need to know the terms of Quadratic Formula, i.e.,
x=[-b±√(b^2-4ac)]/2a
This is applicable for every quadratic equation of the form ax^2+bx+c=0.
Here, in the formula, 'a' is the coefficient of x^2, 'b' is the coefficient of x and 'c' is the remaining constant term.
One thing to be. noted is that before solving any quadratic equation, make sure that the RHS has 0, not any other constant. If it has a constant, take it to the LHS. After doing this, your equation should look something like the general quadratic equation, i.e., ax^2+bx+c=0.
Lets solve all the quadratic equations you've mentioned:
1. 7v^2-48v-7=0:
Here since we have 'v' instead of 'x', there'll be no changes in the formula except for wherever there's 'x', it's to be replaced by 'v'.
Here, a=7, b=-48, c=-7.
Using formula,
v=[-(-48)±√{(-48)^2-4(7)(-7)}]/2(7)
Solving, we get the value of 'v' as 7 and -1/7
2. 9h^2-15h+4=0:
This is equation is in the form of general equation. The only difference is that this equation has 'h' instead of 'x'. So, wherever there's 'x' in quadratic formula, just put 'h'.
Here, a=9, b=-15, c=4
By formula,
h=[-(-15)±√{(-15)^2-4(9)(4)}]/2(9)
Solving, we get the values of 'h' as 4/3 and 1/3.
3. 25x^2+10x+1=0:
The equation is similar to that of quadratic equation, so, no changes.
Here, a=25, b=10, c=1
By formula,
x=[-10±√{(10^2-4(25)(1)}]/2(25)
Solving, we get to see that the roots are real and equal, the value of 'x' is -1/5.
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Paul say: To solve quadratic equations using the formula all you do is take your quadratic equation ax^2 + bx + c = 0 and plug the values for a,b,c into (-b +/- sqrt(b^2 - 4ac))/(2a) into the formula.