Sum of the roots of the equation is 4 x -32 x -3-128-0A- 7B- 8C- 5D- 6

The correct option is A 7 Put 2^x = y Equation becomes y^2 3(8y)+128=0 ⇒ y^2 24y+128=0 ⇒ (y 8)(y 16)=0 ⇒ y =16,8 ⇒ 2^x=16,8 ⇒ x = 4, 3 ∴ Sum of the roots is 7 ...

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Byju's Answer

Standard X

Mathematics

Solving a Quadratic Equation by Completion of Squares Method

Sum of the ro...

Question

Sum of the roots of the equation is $4^{x} - 3 (2^{x + 3}) + 128 = 0$

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Solution

The correct option is C
7

Put $2^{x} = y$ Equation becomes

$y^{2} - 3 (8 y) + 128 = 0 \Rightarrow y^{2} - 24 y + 128 = 0 \Rightarrow (y - 8) (y - 16) = 0 \Rightarrow y = 16, 8 \Rightarrow 2^{x} = 16, 8 \Rightarrow x = 4, 3$

$∴$ Sum of the roots is 7

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Q. Sum of the roots of the equation

4^{x} - 3 (2^{x + 3}) + 128 = 0

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4^{x} - 3 (2^{x + 3}) + 128 = 0

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4^{x} - 3 (2^{x + 3}) + 108 = 0

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Sum of the roots of the equation is 4x−3(2x+3)+128=0

The correct option is C 7 Put 2x=y Equation becomes y2−3(8y)+128=0⇒y2−24y+128=0⇒(y−8)(y−16)=0⇒y=16,8⇒2x=16,8⇒x=4,3 ∴ Sum of the roots is 7

Sum of the roots of the equation is $4^{x} - 3 (2^{x + 3}) + 128 = 0$

The correct option is C
7

Put $2^{x} = y$ Equation becomes

$y^{2} - 3 (8 y) + 128 = 0 \Rightarrow y^{2} - 24 y + 128 = 0 \Rightarrow (y - 8) (y - 16) = 0 \Rightarrow y = 16, 8 \Rightarrow 2^{x} = 16, 8 \Rightarrow x = 4, 3$

$∴$ Sum of the roots is 7