Question

if one of the zeros of the quadratic polynomial 5 x square + kx + 2( K + 1) is 1, then find the value of k.

Answer 1

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}$

If 1 is root of the following equation it will satisfy it.

$5 {x}^{2} + kx + 2(k + 1) = 0 \\ \\ put \: \: x = 1 \\ \\ \implies 5 {(1)}^{2} + k(1) + 2k + 2 = 0 \\ \\ \implies 3k + 7 = 0 \\ \\ \implies \huge \boxed{k = - \frac{7}{3}}$

$\boxed{\begin{minipage}{7cm} </p><p>Something \: Important \: for \; you.. \\ \\ If \: \alpha \: \: and \: \: \beta \: \: are \: \: roots \: \: \\ of \: quadratic \: equation \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \times \beta = \frac{c}{a}</p><p> \end {minipage}}$

Answer 2

Answer:

k=-7/3

Step-by-step explanation:

As 1 is the zero of polynomial

so

5(1)^2+k(1)+2(k+1)=0

5+k+2k+2=0

7+3k=0

k=-7/3