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Sagot :
To solve the quadratic equation [tex]\( 3x^2 - 33x + 54 = 0 \)[/tex], we can follow these steps:
1. Identify the coefficients:
- The quadratic equation is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex].
- Here, [tex]\( a = 3 \)[/tex], [tex]\( b = -33 \)[/tex], and [tex]\( c = 54 \)[/tex].
2. Calculate the discriminant:
- The discriminant ([tex]\( \Delta \)[/tex]) is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
- Substituting the values, we get:
[tex]\[ \Delta = (-33)^2 - 4 \cdot 3 \cdot 54 = 1089 - 648 = 441 \][/tex]
3. Find the roots using the quadratic formula:
- The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Substituting [tex]\( a = 3 \)[/tex], [tex]\( b = -33 \)[/tex], and [tex]\( \Delta = 441 \)[/tex], we get:
[tex]\[ x = \frac{-(-33) \pm \sqrt{441}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{33 \pm 21}{6} \][/tex]
4. Calculate the two solutions:
- First solution [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{33 + 21}{6} = \frac{54}{6} = 9 \][/tex]
- Second solution [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{33 - 21}{6} = \frac{12}{6} = 2 \][/tex]
5. Arrange the solutions from least to greatest:
- The lesser value is 2.
- The greater value is 9.
Therefore, the solutions to the quadratic equation [tex]\( 3x^2 - 33x + 54 = 0 \)[/tex] are:
- Lesser [tex]\( x = 2 \)[/tex]
- Greater [tex]\( x = 9 \)[/tex]
The answers are:
[tex]\[ \text{lesser } x = 2 \][/tex]
[tex]\[ \text{greater } x = 9 \][/tex]
1. Identify the coefficients:
- The quadratic equation is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex].
- Here, [tex]\( a = 3 \)[/tex], [tex]\( b = -33 \)[/tex], and [tex]\( c = 54 \)[/tex].
2. Calculate the discriminant:
- The discriminant ([tex]\( \Delta \)[/tex]) is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
- Substituting the values, we get:
[tex]\[ \Delta = (-33)^2 - 4 \cdot 3 \cdot 54 = 1089 - 648 = 441 \][/tex]
3. Find the roots using the quadratic formula:
- The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Substituting [tex]\( a = 3 \)[/tex], [tex]\( b = -33 \)[/tex], and [tex]\( \Delta = 441 \)[/tex], we get:
[tex]\[ x = \frac{-(-33) \pm \sqrt{441}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{33 \pm 21}{6} \][/tex]
4. Calculate the two solutions:
- First solution [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{33 + 21}{6} = \frac{54}{6} = 9 \][/tex]
- Second solution [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{33 - 21}{6} = \frac{12}{6} = 2 \][/tex]
5. Arrange the solutions from least to greatest:
- The lesser value is 2.
- The greater value is 9.
Therefore, the solutions to the quadratic equation [tex]\( 3x^2 - 33x + 54 = 0 \)[/tex] are:
- Lesser [tex]\( x = 2 \)[/tex]
- Greater [tex]\( x = 9 \)[/tex]
The answers are:
[tex]\[ \text{lesser } x = 2 \][/tex]
[tex]\[ \text{greater } x = 9 \][/tex]
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