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To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for the quadratic equation [tex]\(a x^2 - 7 x + b = 0\)[/tex] given the roots [tex]\(x = \frac{2}{3}\)[/tex] and [tex]\(x = -3\)[/tex], we can use Vieta's formulas which relate the coefficients of a polynomial to sums and products of its roots.
### Step-by-Step Solution:
#### (i) Finding the value of [tex]\(a\)[/tex]:
1. First, recall Vieta's formulas for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots, [tex]\( \alpha + \beta = -\frac{b}{a} \)[/tex]
- The product of the roots, [tex]\( \alpha \beta = \frac{c}{a} \)[/tex]
2. Here, the given quadratic equation is [tex]\( a x^2 - 7 x + b = 0 \)[/tex]. So comparing with the standard form [tex]\( a x^2 + bx + c = 0 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] is [tex]\(-7\)[/tex], so [tex]\( b = -7 \)[/tex]
- Constant term is [tex]\( b \)[/tex], so [tex]\( c = b \)[/tex]
3. The roots are given as [tex]\( \frac{2}{3} \)[/tex] and [tex]\(-3\)[/tex].
4. Using the sum of the roots for the given equation:
[tex]\[ \frac{2}{3} + (-3) = -\frac{b}{a} = \frac{2}{3} - 3 \][/tex]
Simplifying the sum:
[tex]\[ \frac{2}{3} - 3 = \frac{2}{3} - \frac{9}{3} = -\frac{7}{3} \][/tex]
So, we have:
[tex]\[ -\frac{-7}{a} = -\frac{7}{a} \implies a = \frac{7}{\left(\frac{7}{3}\right)} = -3 \][/tex]
#### (ii) Finding the value of [tex]\(b\)[/tex]:
1. Using the product of the roots:
[tex]\[ \left(\frac{2}{3}\right) \times (-3) = \frac{c}{a} = -3 \cdot \frac{2}{3} = -2 \][/tex]
Recall the original polynomial is [tex]\( ax^2 - 7x + b = 0 \)[/tex]. For the product:
[tex]\[ \frac{b}{a} = -2 \][/tex]
Since [tex]\(a = -3\)[/tex]:
[tex]\[ \frac{b}{-3} = -2 \implies b = -2 \cdot -3 = 6 \][/tex]
### Summary:
(i) The value of [tex]\(a\)[/tex] is [tex]\(-3\)[/tex].
(ii) The value of [tex]\(b\)[/tex] is [tex]\(6\)[/tex].
Therefore, the values obtained are:
[tex]\[ a = -3 \quad \text{and} \quad b = 6 \][/tex]
### Step-by-Step Solution:
#### (i) Finding the value of [tex]\(a\)[/tex]:
1. First, recall Vieta's formulas for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]:
- The sum of the roots, [tex]\( \alpha + \beta = -\frac{b}{a} \)[/tex]
- The product of the roots, [tex]\( \alpha \beta = \frac{c}{a} \)[/tex]
2. Here, the given quadratic equation is [tex]\( a x^2 - 7 x + b = 0 \)[/tex]. So comparing with the standard form [tex]\( a x^2 + bx + c = 0 \)[/tex]:
- Coefficient of [tex]\( x \)[/tex] is [tex]\(-7\)[/tex], so [tex]\( b = -7 \)[/tex]
- Constant term is [tex]\( b \)[/tex], so [tex]\( c = b \)[/tex]
3. The roots are given as [tex]\( \frac{2}{3} \)[/tex] and [tex]\(-3\)[/tex].
4. Using the sum of the roots for the given equation:
[tex]\[ \frac{2}{3} + (-3) = -\frac{b}{a} = \frac{2}{3} - 3 \][/tex]
Simplifying the sum:
[tex]\[ \frac{2}{3} - 3 = \frac{2}{3} - \frac{9}{3} = -\frac{7}{3} \][/tex]
So, we have:
[tex]\[ -\frac{-7}{a} = -\frac{7}{a} \implies a = \frac{7}{\left(\frac{7}{3}\right)} = -3 \][/tex]
#### (ii) Finding the value of [tex]\(b\)[/tex]:
1. Using the product of the roots:
[tex]\[ \left(\frac{2}{3}\right) \times (-3) = \frac{c}{a} = -3 \cdot \frac{2}{3} = -2 \][/tex]
Recall the original polynomial is [tex]\( ax^2 - 7x + b = 0 \)[/tex]. For the product:
[tex]\[ \frac{b}{a} = -2 \][/tex]
Since [tex]\(a = -3\)[/tex]:
[tex]\[ \frac{b}{-3} = -2 \implies b = -2 \cdot -3 = 6 \][/tex]
### Summary:
(i) The value of [tex]\(a\)[/tex] is [tex]\(-3\)[/tex].
(ii) The value of [tex]\(b\)[/tex] is [tex]\(6\)[/tex].
Therefore, the values obtained are:
[tex]\[ a = -3 \quad \text{and} \quad b = 6 \][/tex]
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