As per a survey, 75% of students in Australia search for quadratic equation solver or math assignment as they lack a firm grasp of the fundamentals. If you know the right tricks, solving the most intimidating quadratic equation problems can become a lot easier.
Solving Quadratic Equation by Factoring The Equation
As per math experts associated with help with assignment writing services, this is one of the easiest ways to solve quadratic equations. It incorporates the following steps –
- Combine the similar terms and move them to LHS or RHS
Move all of the values to one side of the equation, keeping the x² term positive. Add or subtract all of the x² terms to combine the terms. Move all the x terms and the constants to one side of the equation so that nothing remains on the other side. Once it is done, you can write 0 on that side of the equal sign.
2x²-8x-4=3x-x²
2x²+x²-8x-3x-4=0
3x²-11x-4=0
- Factor the expression
Use the coefficients of the x² term (3) and the factors of the constant term (-4) to factorise the expression. Multiply them and add up to the middle term (-11).
- Set each set of parenthesis as separate equations equal to 0
This step would lead you to find two values for x. It would make the entire equation equal to zero - (3x+1) (x-4) = 0.
- Solve each zeroed equation independently
If you go through algebra tutorials, you would understand that in a quadratic equation, there will always be two possible values for x. You are required to find x for each possible value of x one by one isolating the variable. Then you would have to write down the two solutions for x as the final solution.
Here’s how:
- Solve 5x+1=0
5x = -1….by subtracting
5x/5 = -1/5….by dividing
X = -1/5
- Solve x-6 = 0
X = 6 by subtracting
X= (-1/5, 6) by making a set of possible, separate solutions which signifies x= -1/5, or x = 6
- Check x = -1/5 in (5x+1) (x-6) = 0
Now, we have (5[-1/5] + 1) ([-1/5]-6) = 0
By substituting, we get = (-1+1) (-6 1/5) = 0
By simplifying, we get - (0) (-6 1/5) = 0….by multiplying therefore 0 = 0
Yes, x = -1/5 works.
- Check x = 4 in (3x-1)(x-4)=0
We have (3[4] +1) ([4]-4) x2 = 0
By substituting (13) (4-4) x2 = 0
By simplifying (13) (0) = 0
By multiplying 0 = 0
Therefore, x = 4 works
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