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PE 2 - Quadratic Equation | Algebra - Quadratic Equations

1. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If one root of the equation ax² + bx + c = 0 is four times the other, then ___________

(a)
b² = 25ac
(b)
b² = 4ac
(c)
4b² = 25ac
(d)
None of these

Answer: Option C

Explanation :

Let α, 4α are the roots.

Sum of the roots
α + 4α = $- \frac{b}{a}$
⇒ 5α = $- \frac{b}{a}$
⇒ α = $- \frac{b}{5 a}$ …(1)

Product of the roots
α × 4α = $\frac{c}{a}$
⇒ 4α² = $\frac{c}{a}$
⇒ $\frac{4 b^{2}}{25 a^{2}}$ = $\frac{c}{a}$ [from (1)]

4b² = 25ac.

Hence, option (c).

2. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If f(x) = x² + 4x + 1 and g(x) = x + 3, then the roots of the quadratic equation g[f(x)] will be

(a)
-2, -2
(b)
2, -1
(c)
-5, 4
(d)
None of these

3. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

The quadratic equation g(x) = (px² + qx + r), p ≠ 0, attains its maximum value at x = 5/3. Product of the roots of the equation g(x) is 6. What is the value of q/r?

(a)
-5/3
(b)
-5/9
(c)
0
(d)
Cannot be determined

Answer: Option B

Explanation :

Maximum value of px² + qx + r will occur at x = $- \frac{q}{2 p}$
∴ $- \frac{q}{2 p}$ = $\frac{5}{3}$
⇒ q = $- \frac{10 p}{3}$

Product of the roots = $\frac{r}{p}$ = 6
⇒ r = 6p

∴ $\frac{q}{r}$ = $\frac{- 10 p / 3}{6 p}$ = $- \frac{5}{9}$

Hence, option (b).

4. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If ab + bc + ca = 0 and a, b, c are rational numbers then the roots of the equation (ab + bc – ca)x² + (bc + ca – ab)x + (ac + ab – bc) = 0 are

(a)
rational
(b)
irrational
(c)
non-real
(d)
Cannot be determined

Answer: Option A

Explanation :

The sum of coefficients = (ab + bc – ca) + (bc + ca – ab) + (ca + ab – bc) = ab + bc + ca = 0 (given)

⇒ One of the roots of the given QE is 1. Let the other root be β

∴ Product of the roots = $\frac{a c + a b - b c}{a b + b c - c a}$ = 1 × β

⇒ β = $\frac{a c + a b - b c}{a b + b c - c a}$ , which is rational as a, b, c are rational.

Hence, both the roots are rational.

Hence, option (a).

5. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If the equations x² + 2ax + 3b = 0 and x² + 2bx + 3a = 0, have one root in common, then find the value of (a + b). a ≠ b

(a)
0
(b)
3/4
(c)
-3/4
(d)
Cannot be determined

6. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

The condition that both the roots of quadratic equation ax² + bx + c = 0 are negative is

(a)
a and c have an opposite sign that of b
(b)
b and c have an a opposite sign that of a
(c)
a and b have an opposite sign that of c
(d)
a, b and c are all of same sign

Answer: Option D

Explanation :

For both the roots: (α, ß) to be negative

α + ß < 0 and αß > 0

⇒ $- \frac{b}{a}$ < 0 and $\frac{c}{a}$ > 0

∴ $\frac{b}{a}$ >0 and $\frac{c}{a}$ > 0

i.e., a, b and c are of same sign.

Hence, option (d).

7. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If α and ß are the roots of the equation ax² + bx + c = 0, then find the roots of the equation ab²x² + b²cx + c³ = 0

(a)
α²ß, ß²α
(b)
$- \frac{α^{2} ß}{α + ß}$ , $- \frac{α ß^{2}}{α + ß}$
(c)
None of these
(d)
Cannot be determined

Answer: Option B

Explanation :

Dividing the equation ab²x² + b²cx + c³ = 0 by c², we get

$a {(\frac{b x}{c})}^{2}$ + $b (\frac{b x}{c})$ + c = 0

Since α and ß are roots of ax² + bx + c = 0

∴ $\frac{b x}{c}$ = α or ß

⇒ x = $\frac{c}{b} α$ or $\frac{c}{b} ß$

Now, we know α + ß = -b/a and αß = c/a
Dividing both equations, we get
⇒ $\frac{c}{b}$ = - $\frac{α ß}{α + ß}$

⇒ x = $\frac{c}{b} α$ = $- \frac{α^{2} ß}{α + ß}$ or $\frac{c}{b} ß$ = $- \frac{α ß^{2}}{α + ß}$

Hence, $- \frac{α^{2} ß}{α + ß}$ and $- \frac{α ß^{2}}{α + ß}$ are the roots of the equation ab²x² + b²cx + c³ = 0.

Hence, option (a).

8. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

k is an integer satisfying k² ≤ 30. How many equations of the form x² + kx + 4 = 0 exist such that the roots are real and unequal?

Backspace

7 8 9 4 5 6 1 2 3 0 . -

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9. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

If ${(\frac{x + 1}{x - 3})}^{4}$ - 2 ${(\frac{x + 1}{x - 3})}^{2}$ – 8 = 0, then how many possible values can x have?

Backspace

7 8 9 4 5 6 1 2 3 0 . -

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Answer: 2

Explanation :

Let ${(\frac{x + 1}{x - 3})}^{2}$ = a

The given equation becomes:
a² – 2a – 8 = 0

⇒ (a - 4)(a + 2) = 0
⇒ a = 4 or -2 (rejected since a has to be positive)

⇒ ${(\frac{x + 1}{x - 3})}^{2}$ = 4

Case 1:
$\frac{x + 1}{x - 3}$ = 2
⇒ x + 1 = 2x – 6
⇒ x = 7

Case 2:
$\frac{x + 1}{x - 3}$ = -2
⇒ x + 1 = -2x + 6
⇒ x = 5/3

∴ x can take 2 possible values i.e., 7 and 5/3.

Hence, 2.

10. PE 2 - Quadratic Equation | Algebra - Quadratic Equations

A teacher wrote a quadratic equation of the form x² + bx + c = 0 on the board and asked his students to find the roots. One student copied the coefficient of x incorrectly and found the roots as 2 and 9. Another student copied the constant term incorrectly and found the roots as 4 and 5. Find the correct equation.

(a)
x² + 9x + 18 = 0
(b)
x² – 9x + 18 = 0
(c)
x² + 9x – 18 = 0
(d)
x² – 9x – 18 = 0