IDNLearn.com offers a user-friendly platform for finding and sharing knowledge. Join our knowledgeable community and get detailed, reliable answers to all your questions.
Sagot :
To solve the quadratic equation [tex]\( 15x^2 - 10\sqrt{6}x + 10 = 0 \)[/tex], we need to follow a systematic approach. Here's the step-by-step solution:
1. Identify the coefficients:
- The coefficient [tex]\( a \)[/tex] for [tex]\( x^2 \)[/tex] is 15.
- The coefficient [tex]\( b \)[/tex] for [tex]\( x \)[/tex] is [tex]\(-10\sqrt{6}\)[/tex].
- The constant term [tex]\( c \)[/tex] is 10.
2. Compute the discriminant:
The discriminant of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
- Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-10\sqrt{6})^2 - 4 \times 15 \times 10 \][/tex]
- Calculate [tex]\( (-10\sqrt{6})^2 \)[/tex]:
[tex]\[ (-10\sqrt{6})^2 = 100 \times 6 = 600 \][/tex]
- Calculate [tex]\( 4 \times 15 \times 10 \)[/tex]:
[tex]\[ 4 \times 15 \times 10 = 600 \][/tex]
- Now, find the discriminant:
[tex]\[ \Delta = 600 - 600 = 0 \][/tex]
3. Analyze the discriminant:
Since the discriminant [tex]\( \Delta = 0 \)[/tex], the quadratic equation has exactly one real root (with multiplicity 2). The roots are real and equal.
4. Calculate the root(s):
The formula for the roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Substitute [tex]\( b = -10\sqrt{6} \)[/tex], [tex]\( \Delta = 0 \)[/tex], and [tex]\( a = 15 \)[/tex]:
[tex]\[ x = \frac{-(-10\sqrt{6}) \pm \sqrt{0}}{2 \times 15} \][/tex]
- Simplify the expression:
[tex]\[ x = \frac{10\sqrt{6}}{30} = \frac{\sqrt{6}}{3} \][/tex]
Therefore, the equation has a single root with multiplicity 2:
[tex]\[ x = \frac{\sqrt{6}}{3} \][/tex]
So, for the quadratic equation [tex]\( 15x^2 - 10\sqrt{6}x + 10 = 0 \)[/tex], the discriminant is 0, and the sole root (with multiplicity 2) is:
[tex]\[ \boxed{x = 0.816496580927726} \][/tex]
1. Identify the coefficients:
- The coefficient [tex]\( a \)[/tex] for [tex]\( x^2 \)[/tex] is 15.
- The coefficient [tex]\( b \)[/tex] for [tex]\( x \)[/tex] is [tex]\(-10\sqrt{6}\)[/tex].
- The constant term [tex]\( c \)[/tex] is 10.
2. Compute the discriminant:
The discriminant of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex].
- Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-10\sqrt{6})^2 - 4 \times 15 \times 10 \][/tex]
- Calculate [tex]\( (-10\sqrt{6})^2 \)[/tex]:
[tex]\[ (-10\sqrt{6})^2 = 100 \times 6 = 600 \][/tex]
- Calculate [tex]\( 4 \times 15 \times 10 \)[/tex]:
[tex]\[ 4 \times 15 \times 10 = 600 \][/tex]
- Now, find the discriminant:
[tex]\[ \Delta = 600 - 600 = 0 \][/tex]
3. Analyze the discriminant:
Since the discriminant [tex]\( \Delta = 0 \)[/tex], the quadratic equation has exactly one real root (with multiplicity 2). The roots are real and equal.
4. Calculate the root(s):
The formula for the roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Substitute [tex]\( b = -10\sqrt{6} \)[/tex], [tex]\( \Delta = 0 \)[/tex], and [tex]\( a = 15 \)[/tex]:
[tex]\[ x = \frac{-(-10\sqrt{6}) \pm \sqrt{0}}{2 \times 15} \][/tex]
- Simplify the expression:
[tex]\[ x = \frac{10\sqrt{6}}{30} = \frac{\sqrt{6}}{3} \][/tex]
Therefore, the equation has a single root with multiplicity 2:
[tex]\[ x = \frac{\sqrt{6}}{3} \][/tex]
So, for the quadratic equation [tex]\( 15x^2 - 10\sqrt{6}x + 10 = 0 \)[/tex], the discriminant is 0, and the sole root (with multiplicity 2) is:
[tex]\[ \boxed{x = 0.816496580927726} \][/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.