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