Math / Algebra

QuestionQuestion 6 (a) Form a quadratic equation whose roots are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ . Express your answer in the form $a x^{2}+b x+c=0$ , where $a, b, c$ are integers. (b) Form an equation whose roots are twice those in previous part.

Studdy Solution

STEP 1

1. The roots of the quadratic equation are given as mixed numbers.
2. The standard form of a quadratic equation is $ax^2 + bx + c = 0$ .
3. To form a quadratic equation from its roots, use the relationship between roots and coefficients.

STEP 2

1. Convert the given roots into improper fractions.
2. Use the sum and product of roots to form the quadratic equation.
3. Form a new equation with roots that are twice the original roots.

STEP 3

Convert the mixed numbers to improper fractions. The roots given are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ .
$3 \frac{1}{2} = \frac{7}{2}$ $-2 \frac{1}{3} = -\frac{7}{3}$

STEP 4

Use the sum and product of the roots to form the quadratic equation. If the roots are $r_1$ and $r_2$ , then the quadratic equation is:
$x^2 - (r_1 + r_2)x + r_1r_2 = 0$
Calculate the sum and product of the roots:
Sum: $r_1 + r_2 = \frac{7}{2} + \left(-\frac{7}{3}\right) = \frac{21}{6} - \frac{14}{6} = \frac{7}{6}$
Product: $r_1 \cdot r_2 = \frac{7}{2} \cdot \left(-\frac{7}{3}\right) = -\frac{49}{6}$

STEP 5

Substitute the sum and product into the quadratic equation formula:
$x^2 - \left(\frac{7}{6}\right)x - \frac{49}{6} = 0$
Multiply through by 6 to clear the fractions:
$6x^2 - 7x - 49 = 0$

STEP 6

Form an equation whose roots are twice those in the previous part. If the original roots are $r_1$ and $r_2$ , the new roots are $2r_1$ and $2r_2$ .
Calculate the new sum and product:
New Sum: $2r_1 + 2r_2 = 2(r_1 + r_2) = 2 \times \frac{7}{6} = \frac{7}{3}$
New Product: $(2r_1)(2r_2) = 4(r_1r_2) = 4 \times \left(-\frac{49}{6}\right) = -\frac{196}{6} = -\frac{98}{3}$

STEP 7

Substitute the new sum and product into the quadratic equation formula:
$x^2 - \left(\frac{7}{3}\right)x - \frac{98}{3} = 0$
Multiply through by 3 to clear the fractions:
$3x^2 - 7x - 98 = 0$
The quadratic equations are: (a) $6x^2 - 7x - 49 = 0$ (b) $3x^2 - 7x - 98 = 0$

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QuestionQuestion 6 (a) Form a quadratic equation whose roots are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ . Express your answer in the form $a x^{2}+b x+c=0$ , where $a, b, c$ are integers. (b) Form an equation whose roots are twice those in previous part.

Studdy Solution

STEP 1

1. The roots of the quadratic equation are given as mixed numbers.
2. The standard form of a quadratic equation is $ax^2 + bx + c = 0$ .
3. To form a quadratic equation from its roots, use the relationship between roots and coefficients.

STEP 2

1. Convert the given roots into improper fractions.
2. Use the sum and product of roots to form the quadratic equation.
3. Form a new equation with roots that are twice the original roots.

STEP 3

Convert the mixed numbers to improper fractions. The roots given are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ .
$3 \frac{1}{2} = \frac{7}{2}$ $-2 \frac{1}{3} = -\frac{7}{3}$

STEP 4

STEP 5

Substitute the sum and product into the quadratic equation formula:
$x^2 - \left(\frac{7}{6}\right)x - \frac{49}{6} = 0$
Multiply through by 6 to clear the fractions:
$6x^2 - 7x - 49 = 0$

STEP 6

STEP 7

Substitute the new sum and product into the quadratic equation formula:
$x^2 - \left(\frac{7}{3}\right)x - \frac{98}{3} = 0$
Multiply through by 3 to clear the fractions:
$3x^2 - 7x - 98 = 0$
The quadratic equations are: (a) $6x^2 - 7x - 49 = 0$ (b) $3x^2 - 7x - 98 = 0$

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QuestionQuestion 6 (a) Form a quadratic equation whose roots are 3123 \frac{1}{2}321​ and −213-2 \frac{1}{3}−231​. Express your answer in the form ax2+bx+c=0a x^{2}+b x+c=0ax2+bx+c=0, where a,b,ca, b, ca,b,c are integers. (b) Form an equation whose roots are twice those in previous part.

Studdy Solution

STEP 1

1. The roots of the quadratic equation are given as mixed numbers.2. The standard form of a quadratic equation is ax2+bx+c=0 ax^2 + bx + c = 0 ax2+bx+c=0.3. To form a quadratic equation from its roots, use the relationship between roots and coefficients.

STEP 2

1. Convert the given roots into improper fractions.2. Use the sum and product of roots to form the quadratic equation.3. Form a new equation with roots that are twice the original roots.

STEP 3

Convert the mixed numbers to improper fractions. The roots given are 312 3 \frac{1}{2} 321​ and −213 -2 \frac{1}{3} −231​.312=72 3 \frac{1}{2} = \frac{7}{2} 321​=27​ −213=−73 -2 \frac{1}{3} = -\frac{7}{3} −231​=−37​

STEP 4

STEP 5

Substitute the sum and product into the quadratic equation formula:x2−(76)x−496=0 x^2 - \left(\frac{7}{6}\right)x - \frac{49}{6} = 0 x2−(67​)x−649​=0Multiply through by 6 to clear the fractions:6x2−7x−49=0 6x^2 - 7x - 49 = 0 6x2−7x−49=0

STEP 6

STEP 7

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QuestionQuestion 6 (a) Form a quadratic equation whose roots are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ . Express your answer in the form $a x^{2}+b x+c=0$ , where $a, b, c$ are integers. (b) Form an equation whose roots are twice those in previous part.

1. The roots of the quadratic equation are given as mixed numbers.
2. The standard form of a quadratic equation is $ax^2 + bx + c = 0$ .
3. To form a quadratic equation from its roots, use the relationship between roots and coefficients.

1. Convert the given roots into improper fractions.
2. Use the sum and product of roots to form the quadratic equation.
3. Form a new equation with roots that are twice the original roots.

Convert the mixed numbers to improper fractions. The roots given are $3 \frac{1}{2}$ and $-2 \frac{1}{3}$ .
$3 \frac{1}{2} = \frac{7}{2}$ $-2 \frac{1}{3} = -\frac{7}{3}$

Substitute the sum and product into the quadratic equation formula:
$x^2 - \left(\frac{7}{6}\right)x - \frac{49}{6} = 0$
Multiply through by 6 to clear the fractions:
$6x^2 - 7x - 49 = 0$