Question

All solutions of equation $ 4{{x}^{2}}-40x+51=0 $ lie in the interval.

Answer 1

Hint: Type of question is based on the quadratic equation. As the equation contains the equal sign, which means there will be some particular value of answer, rather than having in the interval form. So we will solve this quadratic equation with the method of middle term splitting.

Complete step by step answer:
Moving ahead with the question in step-wise manner,
We had a quadratic equation i.e. $ 4{{x}^{2}}-40x+51=0 $ , so we had to find the solution of this equation, which means we had to find the value of ‘x’ which will satisfy the particular equation, which are called the roots of equation. So as we know that to find the roots of a quadratic equation we can find it with the splitting middle term method.
Method says that quadratic equation $ a{{x}^{2}}+bx+c=0 $ whose roots are $ \alpha $ and $ \beta $ can be written as $ {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $ , so by comparing we can say that $ -\dfrac{b}{a}=\alpha +\beta $ and $ \dfrac{c}{a}=\alpha \beta $ .
Is for all who have coefficient of $ {{x}^{2}} $ is 1, if it is not one, then we will simply multiply the coefficient of $ {{x}^{2}} $ with the product of roots and rest will remain same. So by comparing this method in our question, we want the sum of roots $ \alpha +\beta $ that will be equal to 40 and product of roots will be, $ \alpha \beta =51\times 4 $ that will be equal to 204.
So now we have to find the value of $ \alpha $ and $ \beta $ such that the sum is 40 and the product is 204. So let us first consider the cases for which we will have product equal to 204; that will be; $ 51\times 4 $ , $ 51\times 2\times 2,17\times 3\times 2\times 2,34\times 6,204\times 1 $ ; as all these cases have product 204. Now from these cases we want that two numbers should have a sum equal to 40. So there is only one such case i.e. $ 34\times 6 $ . So according to splitting method we can write it in equation form as;
 $ \begin{align}
  & 4{{x}^{2}}-40x+51=0 \\
 & 4{{x}^{2}}-(34+6)x+51=0 \\
\end{align} $
Now simplifying it further, by taking the common;
 $ \begin{align}
  & 4{{x}^{2}}-34x-6x+51=0 \\
 & 2x(2x-17)-3(2x-17)=0 \\
\end{align} $
So in above result we can further take $ 2x-17 $ common, so we will get;
 $ (2x-17)(2x-3)=0 $
So from here we can say that, to satisfy the equation either of both term should be zero i.e.;
  $ 2x-17=0 $ or $ 2x-3=0 $
Now on simplifying these two condition to find for what value of x, these condition will be satisfied;
 $ \begin{align}
  & 2x-17=0 \\
 & x=\dfrac{17}{2} \\
\end{align} $ or $ \begin{align}
  & 2x-3=0 \\
 & x=\dfrac{3}{2} \\
\end{align} $
So from here we can say, if ‘x’ will be either of $ \dfrac{17}{2} $ or $ \dfrac{3}{2} $ , then the equation will get zero. Hence the answer is $ \dfrac{17}{2} $ and $ \dfrac{3}{2} $ .

Note: While using the splitting method, look at the coefficient of $ {{x}^{2}} $ it should be one, if it is not then the product of roots will be product of ‘a’ and ‘c’ and sum of root will be ‘b’ where a, b, and, c are the coefficient of quadratic equation $ a{{x}^{2}}+bx+c=0 $ .