PE 2 - Quadratic Equation

Let α, 4α are the roots.

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

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

4b² = 25ac.

Hence, option (c).

g(f(x)) = (x² + 4x + 1) + 3 = x² + 4x + 4 = (x + 2)²

The roots are same i.e., -2

Hence, option (a).

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

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

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

Hence, option (b).

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 = $\frac{ac+ab–bc}{ab+bc–ca}$ = 1 × β

⇒ β = $\frac{ac+ab–bc}{ab+bc–ca}$, which is rational as a, b, c are rational.

Hence, both the roots are rational.

Hence, option (a).

Let the common root of x² + 2ax + 3b = 0 and x² + 2bx + 3a = 0 be α.

∴ α² + 2aα + 3b = 0   …(1) and 
α² + 2bα + 3a = 0   …(2)

⇒ α² + 2aα + 3b = α² + 2bα + 3a
⇒ a(2α - 3) = b(2α - 3)
⇒ (2α - 3)(a – b) = 0
⇒ 2α – 3 = 0
⇒ α = 3/2

Substituting α = 3/2 in the equation (1), we get:
⇒ 9/4 + 3a + 3b = 0 
⇒ a + b = -3/4

Hence, option (c).

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).

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

$a{\left(\frac{bx}{c}\right)}^2$ + $b\left(\frac{bx}{c}\right)$ + c = 0

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

∴ $\frac{bx}{c}$ = α or ß

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

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

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

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

Hence, option (a).

For roots to be real and unequal D > 0

⇒ k² – 4 × 1 × 4 > 0
⇒ k² > 16

Also given in the question that k² < 30

∴ 16 < k² < 30

The only integral values of k satisfying the above inequality is ± 5, i.e., 2 values of k.

∴ 2 equations exist.

Hence, 2.

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

⇒ ${\left(\frac{x+1}{x-3}\right)}^2$ = 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.

Student 1:
Roots are 2 and 9, hence the equation student 1 must have solved is:

x² – (2 + 9)x + 2 × 9 = 0
⇒ x² – 11x + 18 = 0

Since student 1 copied only the coefficient of x incorrectly, he must have copied the constant term correctly.

⇒ Constant term of the original equation is 18.

Student 2:
Roots are 4 and 5, hence the equation student 2 must have solved is:

x²– (4 + 5)x + 4 × 5 = 0
⇒ x² – 9x + 20 = 0

Since student 1 copied only the constant term incorrectly, he must have copied the coefficient of x correctly.

⇒ Coefficient of x in the original equation is -9.

∴ Original equation is x² – 9x + 18 = 0

Hence, option (b).