How do you factor $80{{x}^{2}}+68x+12$ ?
Answer
614.7k+ views
Hint: In this question, we have to find the factors of the given equation. The equation given to us is in the form of a quadratic. Thus, we will apply the discriminant method to solve this problem. We compare the general form of quadratic equation and the given equation to get the value of a, b, and c. Then, we will find the value of discriminant using the formula $D=\sqrt{{{b}^{2}}-4ac}$, and thus find the value of x using the discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .After the necessary calculations, we get two equations for x, thus we solve them separately, which is our required answer.
Complete step by step solution:
According to the question, we have to find the factor of a quadratic equation.
Thus, we will use the discriminant method to get the solution.
The equation is $80{{x}^{2}}+68x+12$ ----------------- (1)
As we know, the general quadratic equation is in form of $a{{y}^{2}}+by+c=0$ ---------- (2)
Thus, on comparing equation (1) and (2), we get $a=80,$ $b=68,$ and $c=12$ ------- (3)
So, now we will find the value of discriminant using the formula $D=\sqrt{{{b}^{2}}-4ac}$ by putting the above values in the formula, we get
$\begin{align}
& \Rightarrow D=\sqrt{{{(68)}^{2}}-4.(80).(12)} \\
& \Rightarrow D=\sqrt{4624-3840} \\
\end{align}$
Thus, on further solving, we get
$\Rightarrow D=\sqrt{784}$
Since D>0 which implies the equation has real roots.
$\Rightarrow D=28$ -------------- (4)
Since we see the discriminant is a real number, thus now we will find the value of x, using the formula,
$\Rightarrow x=\dfrac{-b\pm D}{2a}$
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ --------------- (5)
So, we will put the value of equation (3) and (4) in equation (5), we get
$\Rightarrow x=\dfrac{-\left( 68 \right)\pm 28}{2.\left( 80 \right)}$
On further simplification, we get
$\Rightarrow x=\dfrac{-68\pm 28}{160}$
Therefore, we will split the above equation in terms of (+) and (-), we get
$\Rightarrow x=\dfrac{-68+28}{160}$ -------- (6) , and
$\Rightarrow x=\dfrac{-68-28}{160}$ ---------- (7)
Now, we will first solve equation (6), we get
$\Rightarrow x=\dfrac{-40}{160}$
Therefore, we get
$\Rightarrow x=\dfrac{-1}{4}$
Now we will solve equation (7), we get
$\Rightarrow x=\dfrac{-96}{160}$
Therefore, we get
$\Rightarrow x=\dfrac{-3}{5}$
Therefore, for the quadratic equation $80{{x}^{2}}+68x+12$ , its factors is equal to $\left( x+\dfrac{1}{4} \right)\left( x+\dfrac{3}{5} \right)$
Note: While solving this problem, do step-by-step calculations carefully to avoid mathematical errors to get the correct answer. One of the alternative methods for solving this problem is using the split middle term method or the cross multiplication method to get the solution.
Complete step by step solution:
According to the question, we have to find the factor of a quadratic equation.
Thus, we will use the discriminant method to get the solution.
The equation is $80{{x}^{2}}+68x+12$ ----------------- (1)
As we know, the general quadratic equation is in form of $a{{y}^{2}}+by+c=0$ ---------- (2)
Thus, on comparing equation (1) and (2), we get $a=80,$ $b=68,$ and $c=12$ ------- (3)
So, now we will find the value of discriminant using the formula $D=\sqrt{{{b}^{2}}-4ac}$ by putting the above values in the formula, we get
$\begin{align}
& \Rightarrow D=\sqrt{{{(68)}^{2}}-4.(80).(12)} \\
& \Rightarrow D=\sqrt{4624-3840} \\
\end{align}$
Thus, on further solving, we get
$\Rightarrow D=\sqrt{784}$
Since D>0 which implies the equation has real roots.
$\Rightarrow D=28$ -------------- (4)
Since we see the discriminant is a real number, thus now we will find the value of x, using the formula,
$\Rightarrow x=\dfrac{-b\pm D}{2a}$
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ --------------- (5)
So, we will put the value of equation (3) and (4) in equation (5), we get
$\Rightarrow x=\dfrac{-\left( 68 \right)\pm 28}{2.\left( 80 \right)}$
On further simplification, we get
$\Rightarrow x=\dfrac{-68\pm 28}{160}$
Therefore, we will split the above equation in terms of (+) and (-), we get
$\Rightarrow x=\dfrac{-68+28}{160}$ -------- (6) , and
$\Rightarrow x=\dfrac{-68-28}{160}$ ---------- (7)
Now, we will first solve equation (6), we get
$\Rightarrow x=\dfrac{-40}{160}$
Therefore, we get
$\Rightarrow x=\dfrac{-1}{4}$
Now we will solve equation (7), we get
$\Rightarrow x=\dfrac{-96}{160}$
Therefore, we get
$\Rightarrow x=\dfrac{-3}{5}$
Therefore, for the quadratic equation $80{{x}^{2}}+68x+12$ , its factors is equal to $\left( x+\dfrac{1}{4} \right)\left( x+\dfrac{3}{5} \right)$
Note: While solving this problem, do step-by-step calculations carefully to avoid mathematical errors to get the correct answer. One of the alternative methods for solving this problem is using the split middle term method or the cross multiplication method to get the solution.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

