JEE Mathematics Quadratic Equations MCQs Set 22

Answer: (c) \(x^2-11x+30=0\)

Explanation:
1. Sum \(=5+6=11\), product \(=(\alpha+3)(\beta+3)=6+15+9=30\).
2. The equation is \(x^2-11x+30=0\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(-3,\ -\dfrac{27}4\)

Explanation:
1. Roots of the second equation: \(4\) and \(-1\).
2. \(x=4\): \(32+4k-5=0\) gives \(k=-\dfrac{27}4\); \(x=-1\): \(2-k-5=0\) gives \(k=-3\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) real and unequal

Explanation:
1. Let \(f(x)\) be the left side. \(f(b)=(b-a)(b-c)<0\) (\( b \) lies between a and \( c \)).
2. \(f(x)\to+\infty\) as \(x\to\pm\infty\), so \( f \) changes sign and has two distinct real roots.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) \(p=1,\ q=-2\)

Explanation:
1. \(p+q=-p\) gives \(q=-2p\); \(pq=q\) gives \(q(p-1)=0\).
2. \(q=0\) gives \(p=0\); \(p=1\) gives \(q=-2\). Of the options only \(p=1,q=-2\) fits.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(\dfrac{r-p}{p-q},\ 1\)

Explanation:
1. The coefficients add up to \((p-q)+(q-r)+(r-p)=0\), so \(x=1\) is a root.
2. The product of roots is \(\dfrac{r-p}{p-q}\), so the other root is \(\dfrac{r-p}{p-q}\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(\dfrac{c(a-b)}{a^2}\)

Explanation:
1. \(\alpha\beta(\alpha+\beta+1)=\dfrac ca\left(1-\dfrac ba\right)=\dfrac{c(a-b)}{a^2}\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) None of these

Explanation:
1. Cancelling the common term gives \(x=1\), which is excluded because the equation is undefined at \(x=1\).
2. So there is no root (none of the listed counts).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(\dfrac14\)

Explanation:
1. The roots are \(2\omega\) and \(2\omega^2\) (where \(\omega\) is a cube root of unity), so \(\alpha^3=\beta^3=8\).
2. \(\dfrac18+\dfrac18=\dfrac14\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) \(\dfrac b{ac}\)

Explanation:
1. \(a\alpha+b=-a\beta\) and \(a\beta+b=-a\alpha\).
2. \(-\dfrac{1}{a\beta}-\dfrac1{a\alpha}=-\dfrac{\alpha+\beta}{a\alpha\beta}=\dfrac{b/a}{c}=\dfrac b{ac}\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(-2,\ 2\)

Explanation:
1. Complex roots of a real quadratic occur in conjugate pairs: roots \(1\pm i\).
2. Sum \(=2=-a\), product \(=2=b\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) 2

Explanation:
1. \(D=4(1+3k)^2-28(2k+3)=0\) gives \(9k^2-8k-20=0\).
2. \((k-2)(9k+10)=0\), so \(k=2\) (or \(-\dfrac{10}9\)).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) \(\{3,4,-3,-4\}\)

Explanation:
1. \((x+y)^2=49\) and \((x-y)^2=1\), so \(x+y=\pm7\) and \(x-y=\pm1\).
2. \(x=\pm4\) or \(x=\pm3\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) \(q^2=pr\)

Explanation:
1. First: \(4q^2-4pr\ge0\), so \(q^2\ge pr\). Second: \(4pr-4q^2\ge0\), so \(pr\ge q^2\).
2. Hence \(q^2=pr\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) have negative real parts

Explanation:
1. Sum of roots \(=-\dfrac ba<0\); product \(=\dfrac ca>0\).
2. If real, both are negative; if complex, the real part is \( -b/(2a) \) < 0. So in all cases the real parts are negative.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) rational

Explanation:
1. \(D=(2a+b)^2-8ab=(2a-b)^2\) is a perfect square, so the roots are rational: \(-1\) and \(-\dfrac b{2a}\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(p^2+p=2q\)

Explanation:
1. \(\alpha+\beta=-p\) and \(\alpha^2+\beta^2=p^2-2q\).
2. \(-p=p^2-2q\) gives \(p^2+p=2q\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) \(\dfrac{q-\beta}{\alpha-p}\)

Explanation:
1. Subtracting the equations: \((p-\alpha)x+(q-\beta)=0\), so \(x=\dfrac{q-\beta}{\alpha-p}\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) 40

Explanation:
1. Common root \(r\): adding gives \(2r^2+(a+b)r=0\), so \(r=-\dfrac{a+b}2\).
2. Subtracting gives \((a-b)r=-20\), hence \((a-b)(a+b)=40\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(a<2\)

Explanation:
1. \(D\ge0\) gives \(a\le3\). \(f(3)=a^2-5a+6>0\) gives \(a<2\) or \(a>3\).
2. The vertex \(x=a<3\), so combining gives \(a<2\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) A-2, B-4, C-1, D-3

Explanation:
1. A: \( D = -7, x = (-1 \pm i sqrt7)/4 \).
2. \( B \): \( D = -27, x = (-3 \pm 3 sqrt3 i)/2 \).
3. \( C \): \( x^2 - x + 2 = 0, x = (1 \pm i sqrt7)/2 \).
4. \( D \): \( D = -11, x = (-3 \pm i sqrt11)/2 \).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.