Question

How do you find the solutions of $25{{c}^{2}}+20c=-8c+5{{c}^{2}}$?

Answer 1

This is the question of algebraic expression as it consists of variables, coefficients, constants, and mathematical operations such as addition, subtraction, multiplication and division. In the given question of an expression, you just need to simplify the expression by using mathematical operations and evaluate further. The quadratic formula provides the solution for the quadratic equation:
$a{{x}^{2}}+bx+c=0$. In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows: Roots of the quadratic equation= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Formula used:
The quadratic formula provides the solution for the quadratic equation:
$a{{x}^{2}}+bx+c=0$
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step solution:
Given quadratic equation,
$25{{c}^{2}}+20c=-8c+5{{c}^{2}}$
Simplifying the above quadratic equation, we get
$20{{c}^{2}}+28c=0$
Writing the above equation in a standard form, we get
$20{{c}^{2}}+28c=0$
The quadratic formula provides the solution for the quadratic equation:
$a{{x}^{2}}+bx+c=0$
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Determine the quadratic equation’s coefficients a, b and c:
The coefficient of the given quadratic equation $20{{c}^{2}}+28c=0$ are,
a = 20
b = 28
c = 0
Plug these coefficient into the quadratic formula:
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-28\pm \sqrt{{{\left( 28 \right)}^{2}}-\left( 4\times 20\times 0 \right)}}{2\times 20}$
Solve exponents and square root, we get
$\Rightarrow \dfrac{-28\pm \sqrt{{{\left( 28 \right)}^{2}}-\left( 4\times 20\times 0 \right)}}{2\times 20}$
Performing any multiplication and division given in the formula,
$\Rightarrow \dfrac{-28\pm \sqrt{{{\left( 28 \right)}^{2}}-0}}{2\times 20}$
$\Rightarrow \dfrac{-28\pm \sqrt{{{\left( 28 \right)}^{2}}}}{40}$
$\Rightarrow \dfrac{-28\pm 28}{40}$
We got two values, i.e.
$\Rightarrow \dfrac{-28+28}{40}\ and\ \dfrac{-28-28}{40}$
Solving the above, we get
$\Rightarrow 0\ and\ \dfrac{-56}{40}$
The value of ‘x’= 0
Converting into simplest form, we get
$\Rightarrow \dfrac{-56}{40}=\dfrac{-7}{5}$
$\Rightarrow x=\dfrac{-7}{5}$

Therefore, The possible value of $x$ is 0,$\dfrac{-7}{5}$.

Note: To solve or evaluation these types of expression, we need to know about the:
Solving quadratic equations using the formula
Simplifying radicals
Find prime factors
The general form of quadratic equation is$a{{x}^{2}}+bx+c=0$, where a b and c are the numerical coefficients or constants, and the value of $x$is unknown one fundamental rule is that the value of a, the first constant can never be zero.