If - - are the roots of the equation 4 x -32 x -3-128-0- then -2-2-A- 7B- 24- 25D- 16

The correct option is C 254^x 3(2^x+3) + 128=0 ⇒ 2^2x 24·2^x+128=0 Now put 2^x=y ⇒ y^2 24y+128=0 ⇒ y=16, 8 Now when 2^x=16⇒ x=4 and when 2^x=8⇒ x=3 ∴α^2+β^2=25 ...

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Standard XII

Mathematics

Common Roots

If α, β are t...

Question

If $α, β$ are the roots of the equation $4^{x} - 3 (2^{x + 3}) + 128 = 0$ , then $α^{2} + β^{2} =$

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Solution

The correct option is C $25$
$4^{x} - 3 (2^{x + 3}) + 128 = 0 \Rightarrow 2^{2 x} - 24 \cdot 2^{x} + 128 = 0$
Now put $2^{x} = y$
$\Rightarrow y^{2} - 24 y + 128 = 0 \Rightarrow y = 16, 8$
Now when
$2^{x} = 16 \Rightarrow x = 4$
and when $2^{x} = 8 \Rightarrow x = 3$
$∴ α^{2} + β^{2} = 25$

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If α, β are the roots of the equation 4x−3(2x+3)+128=0, then α2+β2=

The correct option is C 254x−3(2x+3)+128=0⇒22x−24⋅2x+128=0 Now put 2x=y ⇒y2−24y+128=0⇒y=16, 8 Now when 2x=16⇒x=4 and when 2x=8⇒x=3 ∴α2+β2=25

If $α, β$ are the roots of the equation $4^{x} - 3 (2^{x + 3}) + 128 = 0$ , then $α^{2} + β^{2} =$

The correct option is C $25$
$4^{x} - 3 (2^{x + 3}) + 128 = 0 \Rightarrow 2^{2 x} - 24 \cdot 2^{x} + 128 = 0$
Now put $2^{x} = y$
$\Rightarrow y^{2} - 24 y + 128 = 0 \Rightarrow y = 16, 8$
Now when
$2^{x} = 16 \Rightarrow x = 4$
and when $2^{x} = 8 \Rightarrow x = 3$
$∴ α^{2} + β^{2} = 25$