
Can every quadratic equation be solved by factoring or by splitting the middle term? When does one use a quadratic formula?
Answer
476.7k+ views
Hint: Here we have to show whether every quadratic equation can be solved by factoring or by splitting the middle term method and when we use the quadratic formula. So we will define all three methods given then we will give an example of an equation and show whether it is solved by the factorization method or splitting the middle term method or we have to use the quadratic formula.
Complete step by step answer:
From the fundamental theorem of algebra a polynomial with degree $n$ has $n$ solution and we know a quadratic equation has a polynomial with highest degree as $2$ so it will have two solutions.
There are three ways to solve a quadratic equation given below:
1. Factorization- In this method we split the middle term of the equation and try to form a new equation such that the terms are in product form and can be solved further.
2. Quadratic formula- We can use a direct formula for solving quadratic equations of general form $a{{x}^{2}}+bx+c=0$ . The formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and we get two solutions from it.
Let us take an example as follows:
$3{{x}^{2}}+\sqrt{5}x-9=0$
Now if we want to use the factorization method, splitting the middle term is not possible as it is in square root. In this case we use quadratic equations for our solution.
If the roots of the equation or we can say the coefficient of the variable in the equation are rational numbers then using the factorization method is possible but if we switch to irrational numbers or complex numbers the complexity grows and using quadratic formula is preferred.
Hence not every quadratic equation can’t be solved by factoring or by splitting the middle term. In case of irrational or complex roots we use the quadratic formula.
Note:
Polynomial equations are formed by variables and constant along with the equal sign. The exponent of the variable is different for different equations and the higher exponent of the equation is known as the degree of the equation. Quadratic equations can be solved by factoring or quadratic formulas and completing square methods.
Complete step by step answer:
From the fundamental theorem of algebra a polynomial with degree $n$ has $n$ solution and we know a quadratic equation has a polynomial with highest degree as $2$ so it will have two solutions.
There are three ways to solve a quadratic equation given below:
1. Factorization- In this method we split the middle term of the equation and try to form a new equation such that the terms are in product form and can be solved further.
2. Quadratic formula- We can use a direct formula for solving quadratic equations of general form $a{{x}^{2}}+bx+c=0$ . The formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and we get two solutions from it.
Let us take an example as follows:
$3{{x}^{2}}+\sqrt{5}x-9=0$
Now if we want to use the factorization method, splitting the middle term is not possible as it is in square root. In this case we use quadratic equations for our solution.
If the roots of the equation or we can say the coefficient of the variable in the equation are rational numbers then using the factorization method is possible but if we switch to irrational numbers or complex numbers the complexity grows and using quadratic formula is preferred.
Hence not every quadratic equation can’t be solved by factoring or by splitting the middle term. In case of irrational or complex roots we use the quadratic formula.
Note:
Polynomial equations are formed by variables and constant along with the equal sign. The exponent of the variable is different for different equations and the higher exponent of the equation is known as the degree of the equation. Quadratic equations can be solved by factoring or quadratic formulas and completing square methods.
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