Question

Please Solve: [Quadratic equations]​

Answer 1

Please refer the attachment above

Answer 2

x² + 2(m - 1)x + (m + 5) = 0

In the given equation, the coefficient of x² is 1.

Given that the equation has equal and real roots. Thus the given quadrati polynomial is a perfect square.

Consider the given equations which have real and equal roots and where the coefficient of x² is 1.

1.  x² + 2x + 1

2.  x² - 6x + 9

3. x² + 10x + 25

4.  x² - 26x + 169

From these equations we get that "the coefficient of x⁰ (e.g. 4 in x² + 4x + 4) is the square of half the coefficient of x."

From x² + 4x + 4,  we get 4 = (-4/2)².

From x² - 6x + 9,  we get 9 = (-6/2)².

From x² + 10x + 25,  we get 25 = (10/2)².

From x² - 26x + 169,  we get 169 = (-26/2)².

From these examples, we get that,

If  x² + ax + b = 0  has real and equal roots, then  b = (a/2)²

 

According to this concept,

Consider the given equation.

x² + 2(m - 1)x + (m + 5) = 0

Coefficient of x² is 1.

a = 2(m - 1)   ;   b = m + 5

As it has real and equal roots, it is a perfect square, thus,

$\displaystyle \begin{aligned}&b=\left(\frac{a}{2}\right)^2\\ \\ \Longrightarrow\ \ &m+5=\left(\frac{2(m-1)}{2}\right)^2\\ \\ \Longrightarrow\ \ &m+5=(m-1)^2\\ \\ \Longrightarrow\ \ &m+5=m^2-2m+1\\ \\ \Longrightarrow\ \ &m^2-2m+1-m-5=0\\ \\ \Longrightarrow\ \ &m^2-3m-4=0\\ \\ \Longrightarrow\ \ &m^2+m-4m-4=0\\ \\ \Longrightarrow\ \ &m(m+1)-4(m+1)=0\\ \\ \Longrightarrow\ \ &(m+1)(m-4)=0\end{aligned}$

$\\ \\ \\ \therefore\ \ \ \Large m=\bold{-1}\ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \Large m=\bold{4}$

Hence the possible values of m are  -1  and  4.