PE 3 - Quadratic Equation

The roots of the given equation are = $\frac{6\pm \sqrt{36-4c}}{2}$ = 3 ± $\sqrt{9-c}$

Now, 3 + √(9-c) is definitely greater than 1 hence cannot lie between 0 and 1.

∴ The roots which can lie between 0 and 1 will be 3 - $\sqrt{9-c}$.

⇒ 0 < 3 - $\sqrt{9-c}$ < 1

⇒ -3 < - $\sqrt{9-c}$ < -2

⇒ 3 > $\sqrt{9-c}$ > 2

⇒ 9 > 9 – c > 4

⇒ 0 >  -c  > -5

⇒ 0 < c < 5

Hence, option (b).

We know the product of roots in a quadratic equation = c/a 

Since the roots are consecutive natural numbers, hence the product of these consecutive natural numbers = 143.

Let the first root be a – 1 and the other root will be a + 1.

∴ (a - 1)(a + 1) = 143

⇒ a² – 1 = 143

⇒ a² = 144

⇒ a = ± 12 

We will reject a = -12 as a is a natural number.

∴ a = 12

⇒ The roots are 12 – 1 = 11 and 12 + 1 = 13.

∴ Sum of the roots = 11 + 13 = 24 = -k

⇒ k = -24

Hence, option (b).

The roots of the equation 2x² – 5x - 3 = 0 are 3 and -1/2.

Now the roots of the equation ax² + bx + c = 0 are k more than these roots.

Hence, roots of ax² + bx + c = 0 are 3 + k and -1/2 + k.

⇒ x = 3 + k should satisfy the given equation.

⇒ a(3 + k)² + b(3 + k) + c = 0

⇒ ak² + 6ak + 9a + 3b + kb + c = 0

⇒ ak² + 6ak + kb + 9a + 3b + c = 0

⇒ k(ak + 6a + b) + 9a + 3b + c = 0

Hence, option (c).

The amount spent by each boy is Rs. ‘x’
⇒ Total amount spent by boys = 3 × x = 3x

The amount spent by each girl is Rs. ‘x - 20’
⇒ Total amount spent by boys = 2 × (x – 20) = 2(x – 20)

Given, 3x × 2(x - 20) = 750

⇒ x(x – 20) = 125

⇒ x² – 20x – 125 = 0

⇒ (x – 25)(x + 5) = 0

⇒ x = 25 or -5

x has to be positive, hence x = 25.

∴ Amount spent by each girl = 25 – 20 = Rs. 5

Hence, 5.

One of the roots of the given equation = $\frac{2}{\sqrt{7}-\sqrt{5}}$ × $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}$ = √7+√5

Since, a and c are rational, c/a i.e., product of the roots will also be rational.

If x is the other root of this equation, then product of roots, c/a = (√7 + √5) × x

⇒ x = $\frac{c/a}{\sqrt{7}+\sqrt{5}}$ = $\frac{c}{2a}\left(\sqrt{7}-\sqrt{5}\right)$

Now, Sum of the roots = (√7 + √5) + $\frac{c}{2a}\left(\sqrt{7}-\sqrt{5}\right)$ = $-\frac{b}{a}$

⇒ -2b = 2a(√7 + √5) + c(√7 - √5)

Hence, option (d).

The maximum value of the expression $\frac{1}{2x^2+3x+10}$ will occur when 2x² + 3x + 10 is minimum.

2x² + 3x + 10 is minimum at x = -(3/4) = -3/4

∴ Minimum value of 2x² + 3x + 10 = $2{\left(\frac{-3}{4}\right)}^2$ + $3\left(\frac{-3}{4}\right)$ + 10

= $\frac{9}{8}$ - $\frac{9}{4}$ + 10

= $-\frac{9}{8}$ + 10

= $\frac{71}{8}$

∴ The maximum value of $\frac{1}{2x^2+3x+10}$ = $\frac{1}{71/8}$ = $\frac{8}{71}$

Hence, option (d).

Let f(x) = x⁴ - ax³ + bx² + 5x - 6,

and g(x) = x² - 5x + 6
g(x) = (x - 3)(x - 2)

Given, x² - 5x + 6 is a factor of x⁴ - ax³ + bx² + 5x - 6.

It means x² - 5x + 6 is completely divisible by x⁴ - ax³ + bx² + 5x - 6.
∴ x⁴ - ax³ + bx² + 5x - 6 is divisible by (x - 3) as well as (x - 2)

Case 1: x⁴ - ax³ + bx² + 5x - 6 is divisible by (x - 3)
⇒ f(3) = 0
⇒ 3⁴ - a × 3³ + b × 3² + 5 × 3 - 6 = 0
⇒ 27a - 9b = 90
⇒ 3a - b = 10   ...(1)

Case 2: x⁴ - ax³ + bx² + 5x - 6 is divisible by (x - 2)
⇒ f(2) = 0
⇒ 2⁴ - a × 2³ + b × 2² + 5 × 2 - 6 = 0
⇒ 8a - 4b = 20
⇒ 2a - b = 5   ...(2)

Solving (1) and (2), we get
a = b = 5

⇒ a - b = 0

Hence, option (b).

Let a = x + 9 and b = x – 9

∴ The given equation is,

$a^{\frac{1}{3}}$ - $b^{\frac{1}{3}}$ = 3 

Cubing both the sides we get,

⇒ a - b - 3 × $a^{\frac{1}{3}}$ × $b^{\frac{1}{3}}$ × $\left(a^{\frac{1}{3}}-b^{\frac{1}{3}}\right)$ = 27 

⇒ a - b - 3 × ${\left(ab\right)}^{\frac{1}{3}}$× 3 = 27 

⇒ a - b - 9${\left(ab\right)}^{\frac{1}{3}}$ = 27 

⇒ x + 9 - (x - 9) - 9 × ${(x^2-81)}^{\frac{1}{3}}$ = 27

⇒ - 9 × ${(x^2-81)}^{\frac{1}{3}}$ = 9

⇒ ${(x^2-81)}^{\frac{1}{3}}$ = -1

⇒ x² - 81 = -1

⇒ x² = 80

Hence, option (c).

Given, $2^{2x^2-7x+5}$ = 1 = 2⁰

⇒ 2x² - 7x + 5 = 0

⇒ 2x² - 2x - 5x + 5 = 0

 ⇒ (2x - 5)(x - 1) = 0

⇒ x = 5/2 or 1

Hence, option (b).

S = (x − 1)⁴ + 4(x − 1)³ + 6(x − 1 )² + 4(x − 1) + 1

(a + b)⁴ = ⁴C₀ × a⁴ + ⁴C₁ × a³ × b + ⁴C₂ × a² × b² + ⁴C₃ × a × b³ + ⁴C₄ × b⁴

= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

∴ We can see that S = [(x − 1) + 1]⁴ = x⁴

Hence, option (c).