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Formula
Number of solution
Answer
$$2 x ^ { 2 } - 3 x + 1 = 0$$
 2 real roots 
Find the number of solutions
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 }$
 Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times 1$
 Remove negative signs because negative numbers raised to even powers are positive 
$D = 3 ^ { 2 } - 4 \times 2 \times 1$
$D = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times 1$
 Calculate power 
$D = \color{#FF6800}{ 9 } - 4 \times 2 \times 1$
$D = 9 - 4 \times 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
 Multiplying any number by 1 does not change the value 
$D = 9 - 4 \times 2$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 }$
 Multiply $- 4$ and $2$
$D = 9 \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$D = \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
 Subtract $8$ from $9$
$D = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 1 }$
 Since $D>0$ , the number of real root of the following quadratic equation is 2 
 2 real roots 
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