Question

if the difference of the roots of the equation x^2+kx+7=0 is 6 then possible values of k are

Answer 1

Given:

• The difference of roots is 6
• Equation: x² +kx +7 = 0

To find:

The value of k.

Answer:

Let the roots of the given equation be  α and β

So, according to the question

 α - β= 6 ----(1)

Now, comparing the given equation with standard equation

ax² +bx +c =0

$\rightarrow \: a = 1 \\ \rightarrow \: b = k \\ \rightarrow \: c = 7$

We know that,

Sum of zeroes = $\frac{ - coeffcient \: of \: x}{coefficient \: of \: {x}^{2} }$

$= \frac{ - b}{ \: \: a}$

$\implies \alpha + \beta = \frac{ - k}{1} \\ \\ \green{ \implies \: \alpha + \beta = - k \: \: } - - (2)$

Also,

Product of zeroes = $\frac{constant \: term}{coeffcient \: of \: {x}^{2} }$

$= \frac{c}{a}$

$\implies \: \alpha \beta = \frac{7}{1} \\ \\ \green{ \implies \: \alpha \beta = 7} - - (3)$

From equation (1) and (2)

$\alpha - \cancel\beta = 6 \\ \alpha + \cancel\beta =6 - k\\ \\ 2 \alpha =6 - k \\ = > \alpha = \frac{ 6- k}{2}$

Now putting the value of  α in eq(1)

$\alpha - \beta = 6 \\ \\ \implies \frac{6 - k}{2} - \beta = 6 \\ \\ \implies \frac{6 - k}{2} - 6 = \beta \\ \\ \implies \frac{6 - k - 12}{2} = \beta \\ \\ \implies \: \frac{ - k - 6}{2} = \beta \\ \\ \implies \: \beta = \frac{ - k - 6}{2}$

Substituting the values of  α and  β in eq(3)

$\implies \left ( \frac{6 - k}{2} \right) \times \left( \frac{ - k - 6}{2} \right) = 7 \\ \\ \implies \: \left ( \frac{6( - k - 6) - k( - k - 6)}{2 \times 2} \right) = 7 \\ \\ \implies \: \left ( \frac{ - 6k - 36 + {k}^{2} + 6k}{4} \right) = 7 \\ \\ \implies \: \left( { \cancel{ - 6k} \: - 36 + {k}^{2} \cancel {+ 6k}}\right) = 7 \times 4 \\ \\ \implies \: {k}^{2} - 36 = 28 \\ \\ \implies {k}^{2} = 28 + 36 \\ \\ \implies \: {k}^{2} = 64 \\ \\ { \implies \: k = \sqrt{64} } \\ \\ { \red{ \implies \boxed{ \: k = {+8 \: or \: -8}}}}\\ \\ \\ \texttt{Another way} \\ \\ \\ We \:have, \\ \\ \alpha + \beta= -k \\ \alpha - \beta= 6 \\ \alpha \beta = 7 \\Now, \\ \\ (\alpha +\beta)^2-(\alpha-\beta)^2\\ Using \: identities \\ \\(a+b)^2= a^2+b^2+2ab \\(a-b)^2=a^2+b^2-2ab \\ \\ \: \implies \small{(\alpha^2+\beta^2+2\alpha\beta)-(\alpha^2+\beta^2-2\alpha\beta)} \\ \\ = \small{(\alpha^2+\beta^2+2\alpha\beta)-\alpha^2-\beta^2+2\alpha\beta} \\ \\ = \small{\cancel{\alpha^2} \:+\cancel{\beta^2}+2\alpha\beta-\cancel{\alpha^2}\:+\:\cancel{\beta^2}+2\alpha\beta}\\ \\=4\alpha\beta\\ \\ \implies ( \alpha +\beta)^2-(\alpha-\beta)^2 =4\alpha\beta \\ \\ Putting \: values\\ \\(-k)^2-(6)^2=4(7) \\ \\ \implies k^2-36=28 \\ \\ \implies k^2= 28+36 \\ \\ \implies k^2=64\\ \\ \implies k= \sqrt{64} \\ \\ \red{\implies \boxed{k=+8 \: or \: -8}}$