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Sagot :
To find the zeros of the quadratic equation [tex]\( y = 3x^2 - 10x + 5 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 5 \)[/tex].
1. Compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 5 = 100 - 60 = 40 \][/tex]
2. Substitute the discriminant into the quadratic formula:
[tex]\[ x = \frac{-(-10) \pm \sqrt{40}}{2 \cdot 3} = \frac{10 \pm \sqrt{40}}{6} \][/tex]
3. Simplify the square root if possible:
[tex]\[ \sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10} \][/tex]
Therefore:
[tex]\[ x = \frac{10 \pm 2\sqrt{10}}{6} \][/tex]
4. Factor out common terms:
[tex]\[ x = \frac{10}{6} \pm \frac{2\sqrt{10}}{6} = \frac{5}{3} \pm \frac{\sqrt{10}}{3} \][/tex]
5. Combine the terms under the same denominator:
[tex]\[ x = \frac{5 \pm \sqrt{10}}{3} \][/tex]
Therefore, the correct answer is [tex]\( C. \quad x = \frac{5 \pm \sqrt{10}}{3} \)[/tex].
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 5 \)[/tex].
1. Compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 5 = 100 - 60 = 40 \][/tex]
2. Substitute the discriminant into the quadratic formula:
[tex]\[ x = \frac{-(-10) \pm \sqrt{40}}{2 \cdot 3} = \frac{10 \pm \sqrt{40}}{6} \][/tex]
3. Simplify the square root if possible:
[tex]\[ \sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10} \][/tex]
Therefore:
[tex]\[ x = \frac{10 \pm 2\sqrt{10}}{6} \][/tex]
4. Factor out common terms:
[tex]\[ x = \frac{10}{6} \pm \frac{2\sqrt{10}}{6} = \frac{5}{3} \pm \frac{\sqrt{10}}{3} \][/tex]
5. Combine the terms under the same denominator:
[tex]\[ x = \frac{5 \pm \sqrt{10}}{3} \][/tex]
Therefore, the correct answer is [tex]\( C. \quad x = \frac{5 \pm \sqrt{10}}{3} \)[/tex].
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