Solutions: Quadratic Equations - Test 3

Q1. (c) no real roots

$D = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3 < 0$.

Q2. (b) $k < 4$

Real distinct roots $\Rightarrow D > 0$. $16 - 4k > 0 \Rightarrow 16 > 4k \Rightarrow k < 4$.

Q3. (a) $b^2/a$

Equal roots $\Rightarrow (2b)^2 - 4ac = 0 \Rightarrow 4b^2 = 4ac \Rightarrow c = b^2/a$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

$D = 3^2 - 4(1)(4) = 9 - 16 = -7 < 0$. Roots are imaginary.

Q5. $3x^2 - 2x + 1/3 = 0$.
$D = (-2)^2 - 4(3)(1/3) = 4 - 4 = 0$.
Since $D=0$, roots are real and equal. [2]

Q6. $kx^2 - 2kx + 6 = 0$.
For equal roots, $D = 0 \Rightarrow (-2k)^2 - 4(k)(6) = 0$.
$4k^2 - 24k = 0 \Rightarrow 4k(k-6) = 0$.
$k=0$ (rejected as coeff of $x^2$ becomes 0) or $k=6$. k = 6 [2]

Q7. $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$.
Since $D > 0$, roots are real and distinct. [2]

Q8. $D = [-6(p+1)]^2 - 4(p+1)[3(p+9)] = 0$.
$36(p+1)^2 - 12(p+1)(p+9) = 0$. Divide by $12(p+1)$ (since $p \neq -1$):
$3(p+1) - (p+9) = 0 \Rightarrow 3p + 3 - p - 9 = 0 \Rightarrow 2p - 6 = 0 \Rightarrow p = 3$. [3]

Q9. $D = 0 \Rightarrow 4(ac+bd)^2 - 4(a^2+b^2)(c^2+d^2) = 0$.
$a^2c^2 + b^2d^2 + 2abcd - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$.
$2abcd - a^2d^2 - b^2c^2 = 0 \Rightarrow -(ad - bc)^2 = 0 \Rightarrow ad = bc \Rightarrow a/b = c/d$. [3]

Q10. $D = (2mc)^2 - 4(1+m^2)(c^2-a^2) = 0$.
$4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0$.
$m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0$.
$-c^2 + a^2(1+m^2) = 0 \Rightarrow c^2 = a^2(1+m^2)$. [5]

(i) Let breadth $x$. Length $x+3$. Rect Area $x(x+3)$.
Triangle Area $\frac{1}{2} \times x \times 12 = 6x$.
$x(x+3) = 6x + 4 \Rightarrow x^2 + 3x - 6x - 4 = 0 \Rightarrow x^2 - 3x - 4 = 0$. [2]

(ii) $D = (-3)^2 - 4(1)(-4) = 9 + 16 = 25 > 0$. Real distinct roots. [1]

(iii) $(x-4)(x+1) = 0 \Rightarrow x = 4$ (since breadth > 0).
Breadth = 4 m, Length = 7 m. [1]