How do you use the quadratic formula to solve $2.5{{x}^{2}}-2.8x=0.4$ ?
Answer
581.7k+ views
Hint: From the given question we have to find the solutions for the $2.5{{x}^{2}}-2.8x=0.4$ by using the quadratic formula. The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form $a{{x}^{2}}+bx+c=0$, where a, b, and c are coefficients. Then, we plug these coefficients in the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By using this formula, we will get the solutions for the above quadratic equation. First we have to shift the right-hand side part to the left hand side and then we have to proceed further by comparing the equation with the general equation and then we have to apply the quadratic formula.
Complete step by step answer:
From the question given equation is
$\Rightarrow 2.5{{x}^{2}}-2.8x=0.4$
Firstly, we have to shift the right-hand side part of the equation to the left-hand side of the equation.
$\Rightarrow 2.5{{x}^{2}}-2.8x-0.4=0$
Now we have to solve this equation by using quadratic formula as we all know that,
The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form $a{{x}^{2}}+bx+c=0$, where a, b, and c are coefficients. Then, we plug these coefficients in the formula
The formula is
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now the given quadratic equation is in the form of $a{{x}^{2}}+bx+c=0$ then by comparing the given equation with the general quadratic equation we will get the values of a, b and c.
By comparing we will get,
$\Rightarrow a=2.5$
$\Rightarrow b=-2.8$
$\Rightarrow c=-0.4$
Now we have to substitute the above values in their respective positions in the formula.
By substituting the above values in their respective positions in the formula we will get,
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-\left( -2.8 \right)\pm \sqrt{{{\left( -2.8 \right)}^{2}}-4\times 2.5\times -0.4}}{2\times 2.5}$
$\Rightarrow x=\dfrac{2.8\pm \sqrt{7.84+4}}{5}$
$\Rightarrow x=\dfrac{2.8\pm \sqrt{11.84}}{5}$
We can write the approximate value of square root of 11.84 as 3.44,
$\Rightarrow x=\dfrac{2.8\pm 3.44}{5}$
$\Rightarrow x=\dfrac{2.8+3.44}{5}\quad or\quad x=\dfrac{2.8-3.44}{5}$
$\Rightarrow x=\dfrac{6.24}{5}\quad or\quad x=\dfrac{-0.64}{5}$
$\Rightarrow x=1.248\quad or\quad x=-0.128$
Therefore, the solutions of the above equation $2.5{{x}^{2}}-2.8x=0.4$ by using quadratic formula is $x=1.248\quad or\quad x=-0.128$.
Note: Students should know the basic formulas of quadratic equations. In the formula there is a minus sign in front of b students should remember this if they only write simply b then they will get the wrong answer or the entire solution will be wrong. students should be very careful while doing the decimal multiplications and divisions calculation part single decimal point can change the answer value.
Complete step by step answer:
From the question given equation is
$\Rightarrow 2.5{{x}^{2}}-2.8x=0.4$
Firstly, we have to shift the right-hand side part of the equation to the left-hand side of the equation.
$\Rightarrow 2.5{{x}^{2}}-2.8x-0.4=0$
Now we have to solve this equation by using quadratic formula as we all know that,
The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form $a{{x}^{2}}+bx+c=0$, where a, b, and c are coefficients. Then, we plug these coefficients in the formula
The formula is
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now the given quadratic equation is in the form of $a{{x}^{2}}+bx+c=0$ then by comparing the given equation with the general quadratic equation we will get the values of a, b and c.
By comparing we will get,
$\Rightarrow a=2.5$
$\Rightarrow b=-2.8$
$\Rightarrow c=-0.4$
Now we have to substitute the above values in their respective positions in the formula.
By substituting the above values in their respective positions in the formula we will get,
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-\left( -2.8 \right)\pm \sqrt{{{\left( -2.8 \right)}^{2}}-4\times 2.5\times -0.4}}{2\times 2.5}$
$\Rightarrow x=\dfrac{2.8\pm \sqrt{7.84+4}}{5}$
$\Rightarrow x=\dfrac{2.8\pm \sqrt{11.84}}{5}$
We can write the approximate value of square root of 11.84 as 3.44,
$\Rightarrow x=\dfrac{2.8\pm 3.44}{5}$
$\Rightarrow x=\dfrac{2.8+3.44}{5}\quad or\quad x=\dfrac{2.8-3.44}{5}$
$\Rightarrow x=\dfrac{6.24}{5}\quad or\quad x=\dfrac{-0.64}{5}$
$\Rightarrow x=1.248\quad or\quad x=-0.128$
Therefore, the solutions of the above equation $2.5{{x}^{2}}-2.8x=0.4$ by using quadratic formula is $x=1.248\quad or\quad x=-0.128$.
Note: Students should know the basic formulas of quadratic equations. In the formula there is a minus sign in front of b students should remember this if they only write simply b then they will get the wrong answer or the entire solution will be wrong. students should be very careful while doing the decimal multiplications and divisions calculation part single decimal point can change the answer value.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

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

What is the full form of POSCO class 10 social science CBSE

Define Potential, Developed, Stock and Reserved resources

The diagonals of a rhombus are 10cm and 24cm Find the class 10 maths CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

What planets have no moons Which one has only one moon class 10 physics CBSE

