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Number of solution
Answer
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$$x ^ { 2 } - 8 x + 16 = 0$$
$ $ 1 real root (multiple root) $ $
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 16 } = \color{#FF6800}{ 0 }$
$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 16 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 8 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 16$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$D = 8 ^ { 2 } - 4 \times 1 \times 16$
$D = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 16$
$ $ Calculate power $ $
$D = \color{#FF6800}{ 64 } - 4 \times 1 \times 16$
$D = 64 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 16$
$ $ Multiplying any number by 1 does not change the value $ $
$D = 64 - 4 \times 16$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 16 }$
$ $ Multiply $ - 4 $ and $ 16$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 64 }$
$D = \color{#FF6800}{ 64 } \color{#FF6800}{ - } \color{#FF6800}{ 64 }$
$ $ Remove the two numbers if the values are the same and the signs are different $ $
$D = 0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 }$
$ $ Since $ D=0 $ , the number of real root of the following quadratic equation is 1 (multiple root) $ $
$ $ 1 real root (multiple root) $ $
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