Question

if a and B are zeros of x^2-8×+k such that a-B =2 then value of k is​

Answer 1

Given

• p(x)= x²-8x+k
• $\alpha$ and $\beta$ are zeros of p(x)

To Find

• value of k

Solution

• p(x)= x²-8x+k

here a = 1 , b = -8 and c = k

We know that

$Sum~~ of~~ zero = \alpha + \beta = \dfrac{-b}{a}$

$= \alpha + \beta = \dfrac{-( - 8)}{1}$

$= \alpha + \beta = 8$

$Product \: \: of \: \: zero = \alpha \beta = \dfrac{c}{a}$

$\alpha \beta = k$

We know that

• (a-b)² = (a+b)²-4ab

${( \alpha - \beta ) }^{2} ={ ( \alpha + \beta )}^{2} - 4 \alpha \beta$

• putting the values

${( 2) }^{2} ={ (8)}^{2} - 4 k$

$4 =64 - 4 k$

$- 60 = - 4 k$

$k = 15 \bold{ \: }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \hline \bf$

• if a and B are zeros of x^2-8×+k such that a-B =2 then value of k is 15.

More to know

• $\begin{gathered}\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\gray{\begin{gathered}\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\red{ \bigstar} \: \underline{\bf{\orange{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab )\\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}\end{gathered}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\\ \end{gathered}$