Q.If $a, b, c, d$ and $p$ are different real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \leq 0$, then show that $a, b, c$ and $d$ are in G.P.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the difference of the roots of the equation $x^2 - 7x + 10 = 0$ is same as the difference of the roots of the equation $x^2 - 17x + k = 0$, then a divisor of k is (A) $14$ (B) $17$ (C) $6$ (D) $15$

   ›Reveal solutionSolution

   Matching the "difference of roots" of the two quadratics gives $k=70$; among the given options only $14$ divides it.

   Concept and Intuition

   For a quadratic $x^2-(\text{sum})x+(\text{product})=0$, the difference of its two roots can be found without solving for the roots individually, using $(\text{root}_1-\text{root}_2)^2 = (\text{sum})^2-4(\text{product})$ — this follows directly from $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$.

   Step-by-Step Solution

   1. For $x^2-7x+10=0$: roots are $2$ and $5$ (factors as $(x-2)(x-5)$), so the difference of roots is $5-2=3$.
   2. For $x^2-17x+k=0$: sum of roots $=17$, product $=k$. Difference$^2 = 17^2-4k = 289-4k$.
   3. Set the two differences equal (matching magnitudes): $289-4k = 3^2 = 9$.
   4. Solve: $4k = 289-9=280 \Rightarrow k=70$. …

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2. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If the roots of $\sqrt{\dfrac{1-y}{y}} + \sqrt{\dfrac{y}{1-y}} = \dfrac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha+\beta)x^4 - 25\alpha\beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $\gamma$ is (A) $\dfrac{1}{2}$ (B) $4$ (C) $2\pi$ (D) $\sqrt{e + 13}$

   ›Reveal solutionSolution

   Solve the surd equation via a reciprocal substitution to get $\alpha,\beta$, then find which $\gamma$ keeps the resulting biquadratic real-rooted. Answer: $\gamma=1/2$.

   Concept and Intuition

   An equation of the form $\sqrt{u}+\dfrac1{\sqrt u}=k$ (here $u=y/(1-y)$) is best handled by substituting $t=\sqrt u$, turning it into a simple quadratic $t^2-kt+1=0$. Once $\alpha,\beta$ are known, the quartic in $x$ becomes a quadratic in $u=x^2$, and "real roots in $x$" needs a non-negative discriminant AND a non-negative root $u$.

   Step-by-Step Solution

   1. Let $t=\sqrt{y/(1-y)}$, so the equation is $t+1/t=5/2 \Rightarrow 2t^2-5t+2=0 \Rightarrow (2t-1)(t-2)=0 \Rightarrow t=2,\,1/2$.
   2. $y=t^2/(1+t^2)$: for $t=2$, $y=4/5$; for $t=1/2$, $y=1/5$. So $\alpha=1/5,\ \beta=4/5$.
   3. $\alpha+\beta=1$, $\alpha\beta=4/25$, $\beta-\alpha=3/5$.
   4. Quartic: $x^4-4x^2+(\gamma+3/5)=0$. Let $u=x^2$: $u^2-4u+(\gamma+3/5)=0$. …

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3. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If 'a' is a rational number, then the roots of the equation $x^2-3ax+a^2-2a-4=0$ are (A) rational and equal numbers (B) different real numbers (C) different rational numbers only (D) not real numbers

   ›Reveal solutionSolution

   This tests analyzing a discriminant that itself is a quadratic in a parameter; the roots are always real and different, but not always rational — the answer is (B).

   Concept and Intuition

   For $x^2-3ax+(a^2-2a-4)=0$, whether the roots are real/equal/rational depends entirely on the discriminant $D=9a^2-4(a^2-2a-4)$. Since $a$ itself varies over all rationals, we must check whether $D$ is always positive (real & distinct roots guaranteed) and separately whether $D$ is always a perfect square (rational roots guaranteed) — these are two independent questions.

   Step-by-Step Solution

   1. Compute $D = 9a^2-4(a^2-2a-4) = 9a^2-4a^2+8a+16 = 5a^2+8a+16$.
   2. Treat $D$ as a quadratic in $a$: its own discriminant is $8^2-4\cdot5\cdot16 = 64-320=-256<0$, and the leading coefficient $5>0$. So $D(a)>0$ for every real value of $a$ — meaning the roots of the original quadratic are always real and always distinct (never equal), regardless of which rational $a$ is chosen.
   3. Check whether $D$ is always a perfect square for rational $a$: at $a=0$, $D=16=4^2$ (rational roots); but at $a=1$, $D=5+8+16=29$, which is not a perfect square, so $\sqrt{D}$ is irrational and the roots are irrational real numbers. …

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4. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If the values of k for which the equation $x^2+2(k+2)x+6k+7=0$ has equal roots are $k_1$ and $k_2$, then $k_1^2+k_2^2=$ (A) 8 (B) 9 (C) 10 (D) 12

   ›Reveal solutionSolution

   The condition for equal roots is discriminant $=0$; solving the resulting quadratic in $k$ gives $k=3$ and $k=-1$, so $k_1^2+k_2^2=10$.

   Concept and Intuition

   A quadratic $ax^2+bx+c=0$ has equal (repeated) roots exactly when its discriminant $b^2-4ac=0$. Here the coefficients themselves depend on a parameter $k$, so setting the discriminant to zero produces a new quadratic equation whose roots are the two values of $k$ that make the original equation have a double root.

   Step-by-Step Solution

   1. For $x^2+2(k+2)x+(6k+7)=0$: $a=1,\ b=2(k+2),\ c=6k+7$.
   2. Discriminant $=0$: $[2(k+2)]^2-4(1)(6k+7)=0 \Rightarrow 4(k+2)^2-4(6k+7)=0$.
   3. Divide by 4: $(k+2)^2-(6k+7)=0 \Rightarrow k^2+4k+4-6k-7=0 \Rightarrow k^2-2k-3=0$. …

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5. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If the roots of the equation $3x^2+4kx+3=0$ are non-real, then k lies in the interval (A) $\left[-2, \dfrac{-3}{2}\right]$ (B) $\left[\dfrac{3}{2}, 2\right]$ (C) $\left(\dfrac{-3}{2}, \dfrac{3}{2}\right)$ (D) $(2, 3)$

   ›Reveal solutionSolution

   Non-real roots of a quadratic require a strictly negative discriminant; solving that inequality for $k$ gives the open interval $(-3/2, 3/2)$.

   Concept and Intuition

   For $ax^2+bx+c=0$ with real coefficients, the roots are non-real exactly when the discriminant $b^2-4ac<0$ (strictly, since $=0$ gives a repeated real root and $>0$ gives two distinct real roots).

   Step-by-Step Solution

   1. Here $a=3,\ b=4k,\ c=3$.
   2. Discriminant $= (4k)^2-4(3)(3) = 16k^2-36$.
   3. Require $16k^2-36<0 \Rightarrow 16k^2<36 \Rightarrow k^2<\dfrac{36}{16}=\dfrac{9}{4}$.
   4. Taking square roots: $|k|<\dfrac{3}{2}$, i.e. $k\in\left(-\dfrac{3}{2},\dfrac{3}{2}\right)$. …

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6. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If the difference between the roots of $x^2+ax+b=0$ and that of the roots of $x^2+bx+a=0$ is same and $a \neq b$, then (A) $a-b-4=0$ (B) $a-b+4=0$ (C) $a+b+4=0$ (D) $a+b-4=0$

   ›Reveal solutionSolution

   Equating the (squared) root-differences of the two quadratics and factoring out $(a-b)$ (which is nonzero) leaves $a+b+4=0$.

   Concept and Intuition

   For a monic quadratic $x^2+px+q=0$, if the roots are $\alpha,\beta$, then $\alpha-\beta = \pm\sqrt{(\alpha+\beta)^2-4\alpha\beta} = \pm\sqrt{p^2-4q}$ (using $\alpha+\beta=-p,\ \alpha\beta=q$). So the magnitude of the difference of roots depends only on $\sqrt{p^2-4q}$.

   Step-by-Step Solution

   1. For $x^2+ax+b=0$: difference of roots (in magnitude) $=\sqrt{a^2-4b}$.
   2. For $x^2+bx+a=0$: difference of roots (in magnitude) $=\sqrt{b^2-4a}$.
   3. Given these are equal: $\sqrt{a^2-4b} = \sqrt{b^2-4a}$. Squaring (valid since both sides are non-negative real quantities under the given condition): $a^2-4b = b^2-4a$.
   4. Rearranging: $a^2-b^2 -4b+4a = 0 \Rightarrow (a-b)(a+b) + 4(a-b) = 0 \Rightarrow (a-b)(a+b+4)=0$. …

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7. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.For what values of $a \in Z$, the quadratic expression $(x+a)(x+1991)+1$ can be factorised as $(x+b)(x+c)$, where $b,c \in Z$? (A) $1990$ (B) $1989$ (C) $1991$ (D) $1992$

   ›Reveal solutionSolution

   $(x+a)(x+1991)+1$ becomes a perfect square $(x+m)^2$ exactly when $a$ and $1991$ are $2$ apart (symmetric about the same midpoint $m$); among the choices this happens at $a=1989$.

   Concept and Intuition

   For two numbers $p,q$ with mean $m=\frac{p+q}{2}$, we can write $p=m-d$, $q=m+d$ where $d=\frac{q-p}{2}$. Then $(x+p)(x+q) = (x+m-d)(x+m+d) = (x+m)^2-d^2$. Adding $1$ gives $(x+m)^2 - d^2+1$, which becomes a perfect square $(x+m)^2$ precisely when $d^2=1$, i.e. $d=\pm1$ — meaning $p$ and $q$ differ by exactly $2$.

   Step-by-Step Solution

   1. Here $q=1991$ (fixed) and $p=a$ (to determine), so we want $|a-1991|=2$, i.e. $a=1989$ or $a=1993$.
   2. Check $a=1989$: midpoint $m=\frac{1989+1991}{2}=1990$, $d=1$. So $(x+1989)(x+1991)+1=(x+1990)^2-1+1=(x+1990)^2$ — a perfect square with $b=c=1990$, both integers. ✓
   3. Check the other listed options directly using $(b-c)^2=(b+c)^2-4bc$ (must be a non-negative perfect square for integer $b,c$ to exist):
      • $a=1990$: $b+c=3981$, $bc=1990\cdot1991+1$; $(b-c)^2 = (1991-1990)^2-4=1-4=-3 <0$ — no real (hence no integer) solution.
      • $a=1991$: gives $(x+1991)^2+1$, and $(b-c)^2=-4<0$ — no solution. …

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8. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The number of real values of m so that the equation $x^2 + (2m+1)x + m = 0$ has equal roots is (A) 1 (B) 0 (C) 2 (D) 3

   ›Reveal solutionSolution

   Setting the discriminant of the quadratic to zero (the condition for equal/repeated roots) leads to an equation with no real solution, so no real $m$ works.

   Concept and Intuition

   A quadratic $ax^2+bx+c=0$ has equal (repeated) roots exactly when its discriminant $b^2-4ac=0$. Here $a=1,\ b=2m+1,\ c=m$, so we set up and solve this discriminant condition for $m$.

   Step-by-Step Solution

   1. Discriminant: $D=(2m+1)^2-4(1)(m)=4m^2+4m+1-4m=4m^2+1$.
   2. Set $D=0$: $4m^2+1=0\Rightarrow m^2=-\dfrac14$.
   3. Since $m^2$ cannot be negative for real $m$, there is no real value of $m$ satisfying this equation. …

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9. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.If a, b, c, d are real numbers such that $a<b<c<d$, then the roots of the equation $(x-a)(x-c)+2(x-b)(x-d)=0$ are (A) Real & need not be distinct (B) Real and distinct (C) Non-real and distinct (D) Non-real and need not be distinct

   ›Reveal solutionSolution

   This tests root-location by sign analysis (intermediate value theorem) rather than solving the quadratic explicitly. The roots are real and distinct.

   Concept and Intuition

   When a quadratic is built as a sum of two products like $(x-a)(x-c)+k(x-b)(x-d)$ with $k>0$, you don't need to expand and use the discriminant. Instead, evaluate the quadratic at the four given points $a,b,c,d$. Wherever the sign flips between consecutive points, a root must lie strictly in between (continuity + IVT). Since a quadratic has at most 2 roots, finding 2 sign changes locates BOTH roots precisely, and they must be real and distinct because they sit in two separate, non-overlapping open intervals.

   Step-by-Step Solution

   1. Let $f(x)=(x-a)(x-c)+2(x-b)(x-d)$. This is quadratic with leading coefficient $1+2=3>0$.
   2. Evaluate at $x=a$: $f(a)=(a-a)(a-c)+2(a-b)(a-d)=2(a-b)(a-d)$. Since $a<b$ and $a<d$, both factors $(a-b),(a-d)$ are negative, so their product is positive: $f(a)>0$.
   3. Evaluate at $x=b$: $f(b)=(b-a)(b-c)+2(b-b)(b-d)=(b-a)(b-c)$. Here $b-a>0$ but $b-c<0$ (since $b<c$), so $f(b)<0$.
   4. Evaluate at $x=c$: $f(c)=(c-a)(c-c)+2(c-b)(c-d)=2(c-b)(c-d)$. Here $c-b>0$ but $c-d<0$ (since $c<d$), so $f(c)<0$.
   5. Evaluate at $x=d$: $f(d)=(d-a)(d-c)+2(d-b)(d-d)=(d-a)(d-c)$. Both factors positive (since $d$ is the largest), so $f(d)>0$. …

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10. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.If '$a$' is a positive integer such that roots of the equation $7x^{2} - 13x + a = 0$ are rational numbers, then the smallest possible value of '$a$' is ________ (A) 5 (B) 6 (C) 7 (D) 8

    ›Reveal solutionSolution

    Rational roots require a perfect-square discriminant; checking small integers shows $a=6$ is the smallest value that works.

    Concept and Intuition

    For a quadratic with rational (in fact, here, integer) coefficients, the roots are rational exactly when the discriminant is a perfect square (so that $\sqrt{\text{disc}}$ is rational).

    Step-by-Step Solution

    1. Discriminant of $7x^2-13x+a=0$ is $D=169-28a$.
    2. We need $D$ to be a non-negative perfect square.
    3. Check $a=5$: $D=169-140=29$, not a perfect square.
    4. Check $a=6$: $D=169-168=1=1^2$, a perfect square. …

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11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.Which of the following condition imply that roots of the equation $\left(\frac{1}{4}\right)x^2 + bx + c = 0$ are integers? (A) $b^2 - c > 0$ (B) $b\ \&\ c$ are even integers (C) $b^2 - c$ is the square of an integer and $b$ is an integer (D) $b\ \&\ c$ are integers

    ›Reveal solutionSolution

    Clearing the fraction and solving the quadratic shows the roots are $-2b\pm2\sqrt{b^2-c}$;

    these are integers exactly when $b$ is an integer and $b^2-c$ is a perfect square.

    Concept and Intuition

    The leading coefficient $1/4$ is the source of potential trouble: even if $b,c$ are "nice"

    numbers, the quadratic formula's $\pm\sqrt{\text{discriminant}}$ term needs to come out as

    an integer as well, and that requires the discriminant itself (suitably scaled) to be a

    perfect square — merely requiring $b,c$ to be integers (or even integers) is not enough, as

    concrete counterexamples show.

    Step-by-Step Solution

    1. Multiply through by 4: $\tfrac14x^2+bx+c=0 \Rightarrow x^2+4bx+4c=0$.
    2. Quadratic formula: $x = \dfrac{-4b\pm\sqrt{16b^2-16c}}{2} = -2b\pm2\sqrt{b^2-c}$.
    3. For $x$ to be an integer: need $-2b$ to be an integer, and $2\sqrt{b^2-c}$ to be an integer.
    4. If $b$ is an integer, $-2b$ is automatically an integer. If additionally $b^2-c$ is a perfect square (say $k^2$, $k\in\mathbb{Z}$), then $2\sqrt{b^2-c}=2k$ is an integer too — so $x=-2b\pm2k$ is guaranteed to be an integer. This is option (C), and it is sufficient.
    5. Check (B) fails: $b=2$, $c=2$ (both even integers): $b^2-c=4-2=2$, not a perfect square $\Rightarrow$ roots $=-4\pm2\sqrt2$, irrational — counterexample.
    6. Check (D) fails: $b=1$, $c=-1$ (both integers): $b^2-c=1-(-1)=2$, not a perfect square $\Rightarrow$ roots $=-2\pm2\sqrt2$, irrational — counterexample. …

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12. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.Let $m$ and $n$ be two integers such that $0 \le m \le 10$ and $0 \le n \le 10$. Then the number of ordered pairs $(m,n)$ such that $x^2+mx+n=0$ has real roots is ______ (A) 71 (B) 73 (C) 75 (D) 72

    ›Reveal solutionSolution

    Count pairs with $n\le m^2/4$ for each $m=0,\dots,10$, capping $n$ at 10; the total is 73.

    Concept and Intuition

    $x^2+mx+n=0$ has real roots exactly when the discriminant $m^2-4n\ge0$, i.e. $n\le \dfrac{m^2}{4}$. Since $n$ must also stay within $[0,10]$, for each fixed $m$ the number of valid $n$ is $\min(10,\lfloor m^2/4\rfloor)+1$ (the $+1$ counts $n=0$).

    Step-by-Step Solution

    1. For each $m$ from 0 to 10, compute $\lfloor m^2/4\rfloor$ and cap at 10 (since $n\le10$):
       • $m=0$: $0\Rightarrow n=0$: 1 value
       • $m=1$: $0\Rightarrow$ 1 value
       • $m=2$: $1\Rightarrow$ 2 values
       • $m=3$: $2\Rightarrow$ 3 values
       • $m=4$: $4\Rightarrow$ 5 values
       • $m=5$: $6\Rightarrow$ 7 values
       • $m=6$: $9\Rightarrow$ 10 values
       • $m=7$: $12\to$ capped at 10 $\Rightarrow$ 11 values ($n=0..10$) …

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