Question & Answer

Solve the quadratic equation for y: \[\text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}\]

ANSWER Verified Verified
Hint: We will be using the concept of quadratic equation to tackle this question. Formula used for solving quadratic equation of the form \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] is \[\text{( }\!\!\alpha\!\! \text{ , } \!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step-by-step answer:
Before proceeding with the question, we should know about quadratic equations. Quadratic equations are the polynomial equations of degree 2 in one variable of type \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] where a, b, c, \[\in R\] and \[\text{a}\ne \text{0}\]. It is the general form of a quadratic equation where a is called the leading coefficient and c is called the absolute term. The values of x satisfying the quadratic equation are the roots of the quadratic equation \[\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}\].
The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.
Formula for solving quadratic equation of the form \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] is \[\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.....(1)\]
So here in the question the variable is y so the general equation will become
And the given equation is \[\text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}.....\text{(3)}\].
Now comparing equation (2) and equation (3) we get, \[\text{a=5}\], \[\text{b=5}\] and \[\text{c =}\,-10\].
Now substituting the values of a, b and c in equation (1) we get,
 \[\text{y}=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 5\times (-10)}}{2\times 5}.....(4)\]
Now simplifying and rearranging equation (4) we get,
 \[\text{y}=\dfrac{-5\pm \sqrt{25+200}}{10}.....(5)\]
Solving for y in equation (5) we get,
  & \text{y}=\dfrac{-5\pm \sqrt{225}}{10} \\
 & \text{y}=\dfrac{-5\pm 15}{10} \\
We get two values of y which are \[{{\text{y}}_{1}}=\dfrac{-5-15}{10}=\dfrac{-20}{10}=-2\] and \[{{\text{y}}_{2}}=\dfrac{-5+15}{10}=\dfrac{10}{10}=1\].
Hence we get -2 and 1 as the answer.

Note: An alternate method for solving quadratic equations is by factoring the equation.
 \[\,\Rightarrow \text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}.....\text{(1)}\]
Simplifying equation (1) by taking out 5 as common term and then solving for y we get,
  & \,\Rightarrow \,{{\text{y}}^{\text{2}}}\text{+y-2=0} \\
 & \,\Rightarrow {{\text{y}}^{\text{2}}}\text{+2y-y-2=0} \\
 & \,\Rightarrow \text{y(y+2)-1(y+2)=0} \\
 & \,\Rightarrow \text{(y+2) (y-1)=0} \\
 & \,\Rightarrow \text{y=}\,-2,\,+1 \\