
How do you solve $3{{x}^{2}}+14x+15=0$ ?
Answer
556.2k+ views
Hint: We are given a quadratic equation which can be solved by the method of factoring the equation. We shall first find the sum of the roots and the product of the roots of the equation. Further we will find numbers which will add up to the sum of the roots and give the result of their multiplication as the product of their roots.
Complete step-by-step answer:
The factoring method solves the equations in one go. However, this method fails when irrational numbers are present and cannot be grouped or when common factors are not present for grouping. In completing the square method, both sides are required to be made proper squares by adding and/or subtracting numbers accordingly. This method is very time consuming and the calculations become complex as well. Taking the square method mostly leads to solving irrational terms under the roots making the working of the problem more twisted.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $3{{x}^{2}}+14x+15=0$,
We will find numbers by hit and trial whose product is equal to $3\times 15=45$ and whose sum is equal to $14$.
Such two numbers are 9 and 5 as $9+5=14$ and $9\times 5=45$.
Now, factoring the equation:
$\Rightarrow 3{{x}^{2}}+9x+5x+15=0$
Taking common, we get:
$\begin{align}
& \Rightarrow 3x\left( x+3 \right)+5\left( x+3 \right)=0 \\
& \Rightarrow \left( x+3 \right)\left( 3x+5 \right)=0 \\
\end{align}$
Hence, $x+3=0$ or $3x+5=0$
$\Rightarrow x=-3$ or $x=-\dfrac{5}{3}$
Therefore, the roots of the equation are $x=-3,-\dfrac{5}{3}$.
Note:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials. The most efficient method of solving such equations is that of using the well-defined properties of polynomials. The calculations become much simpler and time gets saved.
Complete step-by-step answer:
The factoring method solves the equations in one go. However, this method fails when irrational numbers are present and cannot be grouped or when common factors are not present for grouping. In completing the square method, both sides are required to be made proper squares by adding and/or subtracting numbers accordingly. This method is very time consuming and the calculations become complex as well. Taking the square method mostly leads to solving irrational terms under the roots making the working of the problem more twisted.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $3{{x}^{2}}+14x+15=0$,
We will find numbers by hit and trial whose product is equal to $3\times 15=45$ and whose sum is equal to $14$.
Such two numbers are 9 and 5 as $9+5=14$ and $9\times 5=45$.
Now, factoring the equation:
$\Rightarrow 3{{x}^{2}}+9x+5x+15=0$
Taking common, we get:
$\begin{align}
& \Rightarrow 3x\left( x+3 \right)+5\left( x+3 \right)=0 \\
& \Rightarrow \left( x+3 \right)\left( 3x+5 \right)=0 \\
\end{align}$
Hence, $x+3=0$ or $3x+5=0$
$\Rightarrow x=-3$ or $x=-\dfrac{5}{3}$
Therefore, the roots of the equation are $x=-3,-\dfrac{5}{3}$.
Note:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials. The most efficient method of solving such equations is that of using the well-defined properties of polynomials. The calculations become much simpler and time gets saved.
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