If a-log2- and b-log2- Where - - are the roots of x2-12x-32-0- Then a-b-

The correct option is A 5 We have, #160; x^2 12x+32=0 ⇒ (x 8)(x 4)=0 ⇒ x=8,4 Case 1:Taking α=8 and β =4 a+b=log_28+log_24= ...

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Standard XI

Mathematics

Fundamental Laws of Logarithms

if a=log2α ...

Question

if $a = {log}_{2} α$ and $b = {log}_{2} β$ . Where $α, β$ are the roots of $x^{2} - 12 x + 32 = 0$ . Then $a + b =$

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Solution

The correct option is A $5$
We have,
$x^{2} - 12 x + 32 = 0$
$\Rightarrow (x - 8) (x - 4) = 0$
$\Rightarrow x = 8, 4$

Case-1:
Taking $α = 8$ and $β = 4$

$a + b = {log}_{2} 8 + {log}_{2} 4 = 3 + 2 = 5$

Case-2:
Taking $α = 4$ and $β = 8$

$a + b = {log}_{2} 4 + {log}_{2} 8 = 2 + 3 = 5$

So, $a + b = 5$ .

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if a=log2α and b=log2β. Where α,β are the roots of x2−12x+32=0. Then a+b=

The correct option is A 5We have, x2−12x+32=0⇒(x−8)(x−4)=0⇒x=8,4Case-1:Taking α=8 and β=4a+b=log28+log24=3+2=5Case-2:Taking α=4 and β=8a+b=log24+log28=2+3=5So, a+b=5.

if $a = {log}_{2} α$ and $b = {log}_{2} β$ . Where $α, β$ are the roots of $x^{2} - 12 x + 32 = 0$ . Then $a + b =$