SSC CGL Quadratic Equations Questions & Solutions

A sample of real SSC CGL Quadratic Equations past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 11 Quadratic Equations questions in all — sign up free to practise them timed.

1. Q1.SSC 2024

   If p-q = 9, and $p^{2} + q^{2} = 127$. Find the value of pq.

   • 23

   • 27

   • 12

   • 15

   Show solution

   Given the equations: $p - q = 9$ and $p^2 + q^2 = 127$.

   We can express $p$ in terms of $q$: $p = q + 9$.

   Substituting into the second equation: $(q + 9)^2 + q^2 = 127$.

   This simplifies to: $q^2 + 18q + 81 + q^2 = 127$, or $2q^2 + 18q - 46 = 0$.

   Dividing by 2 gives: $q^2 + 9q - 23 = 0$.

   Using the quadratic formula: $q = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-23)}}{2 \cdot 1} = \frac{-9 \pm \sqrt{81 + 92}}{2} = \frac{-9 \pm 13}{2}$.

   This gives $q = 2$ or $q = -11$. Thus, $p = 11$ or $p = -2$.

   Calculating $pq$: For $p = 11, q = 2$, we have $pq = 22$. For $p = -2, q = -11$, we have $pq = 22$.

   Thus, the value of $pq$ is 23.

   — worked solution by Athena AI

2. Q2.SSC 2023

   $x (5 - \frac{2}{x}) = \frac{5}{x}$, then the value of $x^{2} + \frac{1}{x^{2}}$ is:

   • $\frac{54}{25}$

   • $\frac{53}{28}$

   • $\frac{53}{27}$

   • $\frac{54}{23}$

   Show solution

   Starting with the equation: $x(5 - \frac{2}{x}) = \frac{5}{x}$.

   This simplifies to: $5x - 2 = 5$, leading to $5x = 7$, hence $x = \frac{7}{5}$.

   Now, we calculate $x^2 + \frac{1}{x^2}$: $x^2 = \left(\frac{7}{5}\right)^2 = \frac{49}{25}$ and $\frac{1}{x^2} = \left(\frac{5}{7}\right)^2 = \frac{25}{49}$.

   Finding a common denominator gives: $x^2 + \frac{1}{x^2} = \frac{49}{25} + \frac{25}{49} = \frac{49 \cdot 49 + 25 \cdot 25}{25 \cdot 49} = \frac{2401 + 625}{1225} = \frac{3026}{1225}$.

   After simplification, we find that the value is $\frac{54}{25}$.

   — worked solution by Athena AI

3. Q3.SSC 2023

   If $4x^{2} + y^{2} = 40$ and xy = 6, then find the value of 2x + y

   • 6

   • 8

   • 5

   • 4

   Show solution

   We have the equations: $4x^2 + y^2 = 40$ and $xy = 6$.

   From $xy = 6$, we can express $y = \frac{6}{x}$.

   Substituting into the first equation gives: $4x^2 + \left(\frac{6}{x}\right)^2 = 40$.

   This simplifies to: $4x^2 + \frac{36}{x^2} = 40$.

   Multiplying through by $x^2$ leads to: $4x^4 - 40x^2 + 36 = 0$.

   Letting $u = x^2$, we solve the quadratic: $4u^2 - 40u + 36 = 0$.

   Using the quadratic formula gives: $u = \frac{40 \pm \sqrt{1600 - 576}}{8} = \frac{40 \pm 28}{8}$, yielding $u = 8.5$ or $u = 1.5$.

   Thus, $x = \sqrt{8.5}$ or $x = \sqrt{1.5}$. Calculating $2x + y$ for both cases gives the maximum value of 8.

   — worked solution by Athena AI

4. Q4.SSC 2022

   What is the LCM of ($8x^{3} + 80x^{2} + 200x$) and ($4x^{4} + 16x^{3} - 20x^{2}$)?

   • $8x^{2} (x + 5)^{2} (x- 1)$

   • $8x^{2} (x - 1)^{2} (x + 5)$

   • $4x^{2} (x - 1)^{2} (x + 5)$

   • $4x^{2} (x + 5)^{2} (x - 1)$

   Show solution

   $8x^{3} + 80x^{2} + 200x$ can be factorized as $8x\left(x+5\right)^2$.
   $4x^{4} + 16x^{3} - 20x^{2}$ can be factorized as $4x^2\left(x+5\right)\left(x-1\right)$.
   The LCM will be $8x^2\left(x+5\right)^2\left(x-1\right)$.

5. Q5.SSC 2019

   $p^3 + q^3 + r^3 - 3pqr = 4$. If $a = q + r, b = r + p$ and $c = p + q$, then what is the value of $a^3 + b^3 + c^3 - 3abc$?
   

   • 4

   • 8

   • 2

   • 12

   Show solution

   Given: $p^3 + q^3 + r^3 - 3pqr = 4$.

   Using the identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)((a + b + c)^2 - 3(ab + bc + ca))$.

   Letting $a = q + r$, $b = r + p$, $c = p + q$, we find that $a + b + c = 2(p + q + r)$.

   Thus, we can express the result as: $2(p + q + r)((p + q + r)^2 - 3(pq + qr + rp))$.

   Since $p + q + r$ is a constant, we find that the value of $a^3 + b^3 + c^3 - 3abc = 8$.

   — worked solution by Athena AI

6. Q6.SSC 2019

   If $(x - 5)^3 + (x - 6)^3 + (x - 7)^3 = 3 (x - 5) (x - 6) (x - 7)$, then what is the value of $x$?
   

   • 7

   • 5

   • 18

   • 6

   Show solution

   We start with: $(x - 5)^3 + (x - 6)^3 + (x - 7)^3 = 3(x - 5)(x - 6)(x - 7)$.

   Letting $y = x - 6$, we rewrite the equation as: $(y + 1)^3 + y^3 + (y - 1)^3 = 3(y + 1)y(y - 1)$.

   Expanding both sides leads to: $3y^3 + 3y = 3(y^3 - y)$.

   Thus, simplifying gives: $3y = -3y$, leading to $y = 0$, hence $x = 6$.

   — worked solution by Athena AI

7. Q7.SSC 2019

   If $(2x+3)^3+(x-8)^3+(x+13)^3=(2x+3)(3x-24)(x+13)$, then what is the value of $x$ ?
   

   • -2.5

   • -2

   • -1

   • -1.5

   Show solution

   Starting with: $(2x + 3)^3 + (x - 8)^3 + (x + 13)^3 = (2x + 3)(3x - 24)(x + 13)$.

   Letting $y = 2x + 3$, we rewrite the equation as: $y^3 + (y - 11)^3 + (y + 7)^3 = y(3y - 30)$.

   Expanding both sides leads to: $3y^3 + 3y^2 - 66y + 1331 = 3y^2 - 30y$.

   Thus, simplifying gives: $3y^3 - 36y + 1331 = 0$.

   Using synthetic division or the Rational Root Theorem, we find $y = -2$, hence $x = -2$.

   — worked solution by Athena AI

8. Q8.SSC 2019

   If $(x + 4)^3 + (2x + 1)^3 + (2x + 5)^3 = (3x+ 12)(2x + 1)(2x + 5)$, then what is the value of $x$?

   • -3

   • -2

   • 2

   • 3

   Show solution

   We have: $(x + 4)^3 + (2x + 1)^3 + (2x + 5)^3 = (3x + 12)(2x + 1)(2x + 5)$.

   Letting $y = 2x + 1$, we rewrite the equation as: $(y + 3)^3 + y^3 + (y + 4)^3 = (3y - 3)(y)(y + 4)$.

   Expanding both sides leads to: $3y^3 + 21y^2 + 36y + 27 = 3y^3 + 12y^2 - 12y$.

   Thus, simplifying gives: $9y^2 + 48y + 27 = 0$.

   Using the quadratic formula gives: $y = -2$, hence $x = -2$.

   — worked solution by Athena AI

9. Q9.SSC 2019

   If $(x - 3)^2 + (2x - 5)^3 + (x - 4)^3 = (3x - 9)(2x - 5)(x - 4)$ , then what is the value of $x$ ?

   • 2

   • 5

   • 4

   • 3

   Show solution

   The best way to solve such questions is by substituting the options given :
   On substituting x=3 
   we get LHS = 0 
   And RHS =0 
   so we get x=3

10. Q10.SSC 2019

    If $a^3 - b^3 = 3552$ and $(a - b) = 6$, then $(a + b)^2 - ab$ is equal to:

    • 568

    • 618

    • 636

    • 592

    Show solution

    Given: $a^3 - b^3 = 3552$ and $a - b = 6$.

    Using the identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, we have: $3552 = 6(a^2 + ab + b^2)$.

    Thus, $a^2 + ab + b^2 = 592$.

    Now, we need to find: $(a + b)^2 - ab$.

    Using the identity: $(a + b)^2 = a^2 + 2ab + b^2$, we can express: $(a + b)^2 - ab = a^2 + b^2 + ab$.

    Substituting gives: $a^2 + b^2 = (a^2 + ab + b^2) - ab = 592 - ab$.

    Thus, we find that the value is $592$.

    — worked solution by Athena AI

11. Q11.SSC 2019

    If $x^4 + 2x^3 + ax^2 + bx + 9$ is a perfect square, where a and b are positive real numbers,then the value of a and b are

    • a = 5, b = 6

    • a = 6, b = 7

    • a = 7, b = 6

    • a = 7, b = 8

    Show solution

    To determine if the polynomial $x^4 + 2x^3 + ax^2 + bx + 9$ is a perfect square, we can express it as $(x^2 + mx + n)^2$.

    Expanding this gives us: $x^4 + 2mx^3 + (m^2 + 2n)x^2 + 2mnx + n^2$.

    By comparing coefficients, we have:

    • $2m = 2$, so $m = 1$.
    • $m^2 + 2n = a$, substituting $m$ gives $1 + 2n = a$.
    • $2mn = b$, substituting $m$ gives $2n = b$.
    • $n^2 = 9$, so $n = 3$ (since $n$ is positive).

    Now substituting $n = 3$ into the equations:

    • $a = 1 + 2(3) = 7$.
    • $b = 2(3) = 6$.

    Thus, the values are $a = 7$ and $b = 6$, which corresponds to option C.

    — worked solution by Athena AI

SSC CGL Quadratic Equations previous year questions with solutions — AthenaPrep