Question

find the values of k for which the quadratic equation (k+4)x2+(k+1)x+1=0 has equal roots and also find the roots

Answer 1

Answer:

$k = \bold{5} \ \ \ \ \ \ \ \ \ \ \&\ \ \ \ \ \ \ \ \ \ k = \bold{-3} \\ \\ \\ \underline{\underline{Equations}} \\ \\ 9x^2 + 6x + 1 \ \ \ \ \ \ \ \ \ \ \& \ \ \ \ \ \ \ \ \ \ x^2 - 2x + 1 \\ \\ \\ \underline{\underline{Roots}} \\ \\ 9x^2 + 6x+1\ \ \Longrightarrow \ \bold{-\frac{1}{3}} \\ \\ x^2 - 2x + 1\ \ \Longrightarrow \ \ \ \bold{1}$

Step-by-step explanation:

$If a quadratic equation has equal roots, then discriminant \\ (b^2 - 4ac\ \ \ in \ \ \ ax^2 + bx + c = 0)\ will be zero. \\ \\ \\ Here, \\ \\ (k + 4)x^2 + (k + 1)x + 1 = 0 \\ \\ a = k + 4 \ \ \ ; \ \ \ b = k + 1 \ \ \ ; \ \ \ c = 1$

    $b^2 - 4ac = 0 \\ \\ = (k + 1)^2 - (4 \cdot (k + 4) \cdot 1) = 0 \\ \\ = (k^2 + 2k + 1) - (4k + 16) = 0 \\ \\ = k^2 + 2k + 1 - 4k - 16 = 0 \\ \\ = k^2 - 2k - 15 = 0 \\ \\ \\ So we get another quadratic equation. \\ \\ We can solve this by formula method. But I'm using 'completing the square' method to make it faster.$

$k^2 - 2k - 15 = 0 \\ \\ k^2 - 2k = 15 \\ \\ k^2 - 2k + 1 = 15 + 1 \\ \\ = (k - 1)^2 = 16 \\ \\ \\ k - 1 = \sqrt{16} \\ \\ = k - 1 = \pm 4 \\ \\ \\ k = \pm 4 + 1 \\ \\ = k = 4 + 1 \ \ \ \ \ OR\ \ \ \ \ k = -4 + 1 \\ \\ = k = 5 \ \ \ \ \ \ \ \ \ \ OR\ \ \ \ \ k = -3$

$So we get two values for \ k. \\ \\ \\ \therefore\ (5 + 4)x^2 + (5 + 1)x + 1 = \bold{9x^2 + 6x + 1} \\ \\ \& \\ \\ (-3 + 4)x^2 + (-3 + 1)x + 1 = \bold{x^2 - 2x + 1} \\ \\ are the required quadratic equations.$

$Let's find their roots. \\ \\ \\ As equal roots are had, both \ 9x^2 + 6x + 1\ and \ x^2 - 2x + 1\ are perfect squares.$

$\bold{\underline{\underline{TO\ REMEMBER...}}} \\ \\ \\ If a quadratic equation \ ax^2 + bx + c = 0\ has equal roots, or is a perfect square, then, \\ \\ x = \frac{-b}{2a}$

$In \ 9x^2 + 6x + 1, \\ \\ a = 9\ \ \ ; \ \ \ b = 6 \\ \\ x = \frac{-b}{2a} = \frac{-6}{2 \times 9} = \frac{-6}{18} = \bold{-\frac{1}{3}} \\ \\ \\ \therefore\ Root of \ \ 9x^2 + 6x + 1\ is \ \ \bold{-\frac{1}{3}}. \\ \\ \& \\ \\ 9x^2 + 6x + 1 = (3x + 1)^2$

$In \ \ x^2 - 2x + 1, \\ \\ a = 1\ \ \ ; \ \ \ b = -2 \\ \\ x = \frac{-b}{2a} = \frac{-(-2)}{2 \times 1} = \frac{2}{2} = \bold{1} \\ \\ \\ \therefore\ Root of \ \ x^2 - 2x + 1\ is \ \ \bold{1} \\ \\ \& \\ \\ x^2 - 2x + 1 = (x - 1)^2$

$Hope this may be helpful. \\ \\ Please mark my answer as the \ \bold{brainliest}\ if this may be helpful. \\ \\ Thank you. Have a nice day. \\ \\ \\ \#adithyasajeevan$