Answer
396.9k+ views
Hint: As the given equation is a quadratic equation in one variable, we will use the quadratic formula to find the factors of the given equation. If the given quadratic equation is of the form $a{{x}^{2}}+bx+c=0$ the quadratic formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Complete step-by-step solution:
We have been given an equation $5{{x}^{2}}+13x-6$.
We have to find the factors of the given equation.
First we will compare the given equation with the standard quadratic equation which is given by $a{{x}^{2}}+bx+c=0$.
On comparing we get the values $a=5,b=13\And c=-6$
Now, we know that the quadratic formula is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values in the above formula we get
$\Rightarrow x=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\times 5\times -6}}{2\times 5}$
Now, on solving the obtained equation we get
$\begin{align}
& \Rightarrow x=\dfrac{-13\pm \sqrt{169+120}}{10} \\
& \Rightarrow x=\dfrac{-13\pm \sqrt{289}}{10} \\
\end{align}$
Now, we know that the value of square root $\sqrt{289}=17$
Now, substituting the value in the above equation and solving further we get
$\Rightarrow x=\dfrac{-13\pm 17}{10}$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$\Rightarrow x=\dfrac{-13+17}{10}and\Rightarrow x=\dfrac{-13-17}{10}$
Now, let us first consider
$\Rightarrow x=\dfrac{-13+17}{10}$
On solving we get
$\begin{align}
& \Rightarrow x=\dfrac{4}{10} \\
& \Rightarrow x=\dfrac{2}{5} \\
\end{align}$
Now, let us consider $\Rightarrow x=\dfrac{-13-17}{10}$
On solving we get
$\begin{align}
& \Rightarrow x=\dfrac{-30}{10} \\
& \Rightarrow x=-3 \\
\end{align}$
So the two factors of the equation $5{{x}^{2}}+13x-6$ will be $\left( 5x-3 \right)\left( x-2 \right)$.
Note: Avoid calculation mistakes because single calculation mistakes lead to the incorrect answer. To solve a quadratic equation students can use factorization method, completing the square method or quadratic formula method. When the time is less and we are sure about the quadratic formula, then it is best to use this method. We can cross verify the factors by opening the parenthesis and solving.
Complete step-by-step solution:
We have been given an equation $5{{x}^{2}}+13x-6$.
We have to find the factors of the given equation.
First we will compare the given equation with the standard quadratic equation which is given by $a{{x}^{2}}+bx+c=0$.
On comparing we get the values $a=5,b=13\And c=-6$
Now, we know that the quadratic formula is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values in the above formula we get
$\Rightarrow x=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\times 5\times -6}}{2\times 5}$
Now, on solving the obtained equation we get
$\begin{align}
& \Rightarrow x=\dfrac{-13\pm \sqrt{169+120}}{10} \\
& \Rightarrow x=\dfrac{-13\pm \sqrt{289}}{10} \\
\end{align}$
Now, we know that the value of square root $\sqrt{289}=17$
Now, substituting the value in the above equation and solving further we get
$\Rightarrow x=\dfrac{-13\pm 17}{10}$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$\Rightarrow x=\dfrac{-13+17}{10}and\Rightarrow x=\dfrac{-13-17}{10}$
Now, let us first consider
$\Rightarrow x=\dfrac{-13+17}{10}$
On solving we get
$\begin{align}
& \Rightarrow x=\dfrac{4}{10} \\
& \Rightarrow x=\dfrac{2}{5} \\
\end{align}$
Now, let us consider $\Rightarrow x=\dfrac{-13-17}{10}$
On solving we get
$\begin{align}
& \Rightarrow x=\dfrac{-30}{10} \\
& \Rightarrow x=-3 \\
\end{align}$
So the two factors of the equation $5{{x}^{2}}+13x-6$ will be $\left( 5x-3 \right)\left( x-2 \right)$.
Note: Avoid calculation mistakes because single calculation mistakes lead to the incorrect answer. To solve a quadratic equation students can use factorization method, completing the square method or quadratic formula method. When the time is less and we are sure about the quadratic formula, then it is best to use this method. We can cross verify the factors by opening the parenthesis and solving.
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