# CRE 3 - Forming a Quadratic Equation and Relation between roots and coefficients | Algebra - Quadratic Equations

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**CRE 3 - Forming a Quadratic Equation and Relation between roots and coefficients | Algebra - Quadratic Equations**

Form a quadratic equation whose roots at 5 and 6.

- (a)
x

^{2}– 11x - 30 = 0 - (b)
x

^{2}– 11x + 30 = 0 - (c)
x

^{2}+ 11x + 30 = 0 - (d)
None of these

Answer: Option B

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**Explanation** :

We know a quadratic equation whose roots are α and β can be written as (x - α)(x - β) = 0.

∴ The required quadratic equation is (x - 5)(x – 6) = 0

⇒ x^{2} – 5x – 6x + 30 = 0

⇒ x^{2} – 11x + 30 = 0

Hence, option (b).

Workspace:

**CRE 3 - Forming a Quadratic Equation and Relation between roots and coefficients | Algebra - Quadratic Equations**

Form a quadractic equation whose sum of the roots is 2 and product is -15.

- (a)
x

^{2}– 2x + 15 = 0 - (b)
x

^{2}+ 2x – 15 = 0 - (c)
x

^{2}– 2x – 15 = 0 - (d)
None of these

Answer: Option C

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**Explanation** :

We know a quadratic equation can be written as x^{2} – (sum of the roots)x + (product of the roots) = 0

∴ The required quadratic equation is:

⇒ x^{2} – 2x – 15 = 0

Hence, option (b).

Workspace:

If α and β are the roots of the equation 2x^{2} - 3x - 5 = 0, then find the value of 1/α + 1/β?

- (a)
-3/5

- (b)
-2/3

- (c)
3/5

- (d)
None of these

Answer: Option A

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**Explanation** :

Given that α and β are the roots of the equation 2x^{2} - 3x - 5 = 0.

∴ α + β = $-\frac{-3}{2}$ = $\frac{3}{2}$ and α × β = ${\frac{-5}{2}}_{}$

Now, $\frac{1}{\alpha}$+ $\frac{1}{\beta}$ = $\frac{\alpha +\beta}{\alpha \beta}$ = $\frac{3/2}{-5/2}$ = $-\frac{3}{5}$

Hence, option (a).

Workspace:

If α and ß are the roots of the equation x^{2} – 3x + 5 = 0, then what is the value of α^{3} + ß^{3}?

- (a)
18

- (b)
-18

- (c)
-10

- (d)
10

Answer: Option B

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**Explanation** :

We know, (α + ß)^{3} = α^{3} + ß^{3} + 3αß(α + ß)

⇒ α^{3} + ß^{3} = (α + ß)^{3} - 3αß(α + ß) …(1)

Here, α + ß = sum of the roots = - (-3/1) = 3

and, αß = product of the roots = 5/1 = 5.

Substituting these values in (1), we get

α^{3} + ß^{3} = (3)^{3} – 3 × 5 × (3)

⇒ α^{3} + ß^{3} = 27 – 45 = -18

Hence, option (b).

Workspace:

The roots of the quadratic equation 9x^{2} - 36x + a = 0 are in the ratio 2 : 1. Find the value of a.

- (a)
16

- (b)
32

- (c)
32/9

- (d)
None of these

Answer: Option B

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**Explanation** :

The roots of the equatino 9x^{2} - 36x + a = 0 are in the ratio 2 : 1.

Let the roots be α and 2α.

⇒ Sum of the roots = -b/a

⇒ α + 2α = -(-36)/9 = 4

⇒ 3α = 4

⇒ α = 4/3

Now, product of the roots = c/a

⇒ α × 2α = a/9

⇒ 18α^{2} = a

⇒ a = 18 × 16/9 = 32.

Hence, option (b).

Workspace:

Lomror & Rohit attempted to solve a quadratic equation. Lomror made a mistake while noting down the constant term and got the roots as 4 & 3. Rohit made a mistake while noting down the coefficient of x and got the roots as 2 & 3. Find the roots of the actual quadratic equation.

- (a)
-1 and -6

- (b)
1 and -6

- (c)
1 and 6

- (d)
None of these

Answer: Option C

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**Explanation** :

We know, in a quadratic equation ax^{2} + bx + c = 0,

Sum of the roots = -b/a and product of the roots = c/a

Lomror wrote the contant term wrong but he wrote the coefficient of x correctly, hence the sum of the roots he got would be same as the sum of the roots of the actual equation.

∴ Sum of the roots of the actual equation = 4 + 3 = 7

Rohit wrote the coefficient of x wrong but he wrote the constsant term correctly, hence the product of the roots he got would be same as the product of the roots of the actual equation.

∴ Product of the roots of the actual equation = 2 × 3 = 6

Hence, the original equation is x^{2} - 7x + 6 = 0.

⇒ (x - 6)(x - 1) = 0

⇒ x = 1 or 6.

Hence, option (c).

Workspace:

Find the quadratic equation whose one root is 3 + 2√2 and has rational coefficients.

- (a)
x

^{2}+ 6x + 1 = 0 - (b)
x

^{2}- 6x - 1 = 0 - (c)
x

^{2}- 6x + 1 = 0 - (d)
None of these

Answer: Option C

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**Explanation** :

We know, in a quadratic equation with rational coefficients if one root is a + √b, the other root will be a - √b.

Hence, if one the roots of the given quadratic equation is 3 + 2√2, the other root will be 3 - 2√2.

∴ Sum of the roots = 3 + 2√2 + 3 - 2√2 = 6, and

Product of the roots = (3 + 2√2) × (3 - 2√2) = 9 - 8 = 1

∴ The quadratic equation is x^{2} - (Sum)x + Product = 0

⇒ x^{2} - 6x + 1 = 0

Hence, option (c).

Workspace:

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