Quadratic Equations - Test Papers

  CBSE Test Paper 01

Chapter 4 Quadratic Equation

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1. (x+1)2−x2=0$(x+1{)}^2-x^2=0$ has (1)
   1. no real roots
   2. 1 real root
   3. 2 real roots
   4. 4 real roots
2. 9x2 + 12x + 4 = 0 have (1)
   1. Real and Distinct roots
   2. No real roots
   3. Distinct roots
   4. Real and Equal roots
3. If the equation (a2+b2)x2−2(ac+bd)x+c2+d2=0$(a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0$ has equal roots, then (1)
   1. ad=bc$ad=bc$
   2. ab=cd$ab=cd$
   3. ad=bc−−√$ad=\sqrt{bc}$
   4. ab=cd−−√$ab=\sqrt{cd}$
4. The ratio of sum and the product of the roots of 7x2−12x+18=0$7x^2-12x+18=0$ is (1)
   1. 2:3$2:3$
   2. 3:2$3:2$
   3. 7:18$7:18$
   4. 7:12$7:12$
5. If y = 1 is the common root of ly2+ly+3=0$l{y}^2+ly+3=0$ and y2+y+m=0$y^2+y+m=0$, then the value of ‘lm$lm$’ is (1)
   1. 3
   2. – 4
   3. 4
   4. – 3
6. Solve the quadratic equations by factorization method: x2 - 9 = 0 (1)
7. Find the values of p for which the quadratic equation 4x2 + px + 3 = 0 has equal roots. (1)
8. Form a quadratic equation whose roots are -3 and 4. (1)
9. If x = −12$\frac{-1}{2}$ is a solution of the quadratic equation 3x2 + 2kx + 3 = 0, find the value of k. (1)
10. Write the discriminant of the given quadratic equation x2 + x - 12 = 0 (1)
11. Find the values of k for which the given equation has real and equal roots: (k + 1)x2 - 2(k - 1)x + 1 = 0 (2)
12. Check, whether the quadratic equation have real roots and if so, then find the roots of equation. 6x2 + x - 2 = 0 (2)
13. Check whether the given equation is quadratic equation: (x-3) (2x + 1) = x(x + 5) (2)
14. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. (3)
15. If 2 is a root of the quadratic equation 3x2 + px - 8 = 0 and the quadratic equation 4x2- 2px + k = 0 has equal roots, find k. (3)
16. If p, q, r and s are real numbers such that pr = 2(q + s), then show that at least one of the equations x2 + px + q = 0 and x2 + rx + s = 0 has real roots. (3)
17. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream. (3)
18. A train travelling at a uniform speed for 360 km,would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train. (4)
19. Solve for x: 3–√x2+10x+73–√=0$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$ (4)
20. Solve for x: 2(x+22x−3$\frac{x+2}{2x-3}$) - 9(2x−3x+2$\frac{2x-3}{x+2}$) = 3; given that x≠$\ne$ -2, x≠$\ne$32$\frac{3}{2}$ (4)

CBSE Test Paper 01
Chapter 4 Quadratic Equation

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Solution

1. 2. 1 real root
      Explanation: Given: (x+1)2−x2=0$(x+1{)}^2-x^2=0$
      ⇒$\Rightarrow$x2+1+2x−x2=0$x^2+1+2x-x^2=0$
      ⇒$\Rightarrow$2x+1=0$2x+1=0$
      ⇒$\Rightarrow$x=−12$x=\frac{-1}{2}$
      Therefore, (x2+1)2−x2=0$(x^2+1{)}^2-x^2=0$ is a linear polynomial and has one real root.
2. 4. Real and Equal roots
      Explanation: Comparing the given equation to the below equation
      ax2 + bx + c = 0
      a = 9, b = 12, c = 4
      D = b2 - 4ac
      D = 122 - 4×$\times$9×$\times$4
      D = 144 - 144
      D = 0
      If b2−4ac=0 then equation have equal and real roots.
3. 1. ad = bc
      Explanation If the equation (a2+b2)x2−2(ac+bd)x+c2+d2=0$(a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0$ has equal roots, then
      b2−4ac=0$b^2-4ac=0$
      ⇒$\Rightarrow$[−2(ac+bd)]2−4×(a2+b2)×(c2+d2)=0${\left[-2\left(ac+bd\right)\right]}^2-4\times \left(a^2+b^2\right)\times \left(c^2+d^2\right)=0$
      ⇒$\Rightarrow$4[a2c2+b2d2+2abcd]−4[a2c2+a2d2+b2c2+b2d2]=0$4\left[a^2{c}^2+b^2{d}^2+2abcd\right]-4\left[a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\right]=0$
      ⇒$\Rightarrow$4[a2c2+b2d2+2abcd−a2c2−a2d2−b2c2−b2d2]=0$4\left[a^2{c}^2+b^2{d}^2+2abcd-a^2{c}^2-a^2{d}^2-b^2{c}^2-b^2{d}^2\right]=0$
      ⇒$\Rightarrow$a2d2+b2c2−2abcd=0$a^2{d}^2+b^2{c}^2-2abcd=0$
      (ad-bc)2 =0
      ⇒$\Rightarrow$(ad−bc)2=0${\left(ad-bc\right)}^2=0$
      ⇒$\Rightarrow$ad−bc=0$ad-bc=0$
      ⇒$\Rightarrow$ad=bc$ad=bc$
4. 1. 2:3
      Explanation: Ratio of sum and product of the roots of 7x2−12x+18=0$7x^2-12x+18=0$ is α+βαβ$\frac{\alpha +\beta }{\alpha \beta }$
      ⇒$\Rightarrow$
      ⇒$\Rightarrow$−bc$\frac{-b}{c}$
      ⇒$\Rightarrow$1218=23$\frac{12}{18}=\frac{2}{3}$ = 2 : 3
5. 1. 3
      Explanation: In quadratic equation ly2+ly+3=0$l{y}^2+ly+3=0$,
      l(1)2+l(1)+3=0$l{\left(1\right)}^2+l\left(1\right)+3=0$
      ⇒$\Rightarrow$l+l+3=0$l+l+3=0$
      ⇒$\Rightarrow$2l+3=0$2l+3=0$
      ⇒$\Rightarrow$l=−32$l=\frac{-3}{2}$
      And (1)2+1+m=0${\left(1\right)}^2+1+m=0$
      ⇒$\Rightarrow$1+1+m=0$1+1+m=0$
      ⇒$\Rightarrow$2+m=0$2+m=0$
      ⇒$\Rightarrow$m=−2$m=-2$
      ∴$\therefore$lm=−32×(−2)$lm=\frac{-3}{2}\times \left(-2\right)$ = 3

6. We have,
   x2−9=0$x^2-9=0$
   ⇒$\Rightarrow$ (x−3)(x+3)=0$(x-3)(x+3)=0$
   ⇒$\Rightarrow$ x−3=0$x-3=0$ or, x+3=0$x+3=0$
   ⇒$\Rightarrow$ x = 3 or, x = -3 ⇒$\Rightarrow$ x = ± 3
   Thus, x = 3 and x = - 3 are roots of the given equation.

7. 4x2+px+3=0$4x^2+px+3=0$
   a=4,b=p$a=4,b=p$ and c=3$c=3$
   As the equation has equal roots
   ∴$\therefore$ D=0$D=0$
   D=b2−4ac=0$D=b^2-4ac=0$
   or, p2−4×4×3=0$p^2-4\times 4\times 3=0$
   or, p2−48=0$p^2-48=0$
   or, p2=48$p^2=48$
   or, p=±43–√$p=\pm 4\sqrt{3}$

8. We have, x = 4 and x = -3.
   Then,
   x - 4 = 0 and x + 3 = 0
   ⇒$\Rightarrow$ (x - 4)(x + 3) = 0
   ⇒$\Rightarrow$ x2 + 3x - 4x - 12 = 0
   ⇒$\Rightarrow$ x2 - x - 12 = 0
   This is the required quadratic equation

9. we have, 3x2+2kx+3=0$3x^2+2kx+3=0$
   put, x = −12$\frac{-1}{2}$ (given)
   ⇒$\Rightarrow$ 3(−12$\frac{-1}{2}$)2 + 2k$2k$(−12$\frac{-1}{2}$)+3=0$+3=0$
   ⇒$\Rightarrow$ 3(14$\frac{1}{4}$) - k + 3 = 0
   ⇒$\Rightarrow$ 34$\frac{3}{4}$ - k +3 = 0
   ⇒$\Rightarrow$ k=3+$k=3+$ 34$\frac{3}{4}$
   ∴$\therefore$ k=$k=$ 154$\frac{15}{4}$

10. The given quadratic equation is x2+x−12=0,$x^2+x-12=0,$
    here a=1, b=1, c=-12
    ∴D=b2−4ac=(1)2−4((1)(−12)=1+48=49$\therefore D=b^2-4ac=(1{)}^2-4((1)(-12)=1+48=49$
    Hence, the discriminant is 49.

11. We have, (k+1)x2 - 2(k - 1)x+1 = 0.
    a = k + 1, b = -2(k - 1), c = 1.
    D = b2 - 4ac =4(k-1)2 - 4(k + 1) =4(k2 -3k)
    The given equation will have real and equal roots, if
    D = 0 ⇒$\Rightarrow$ 4 (k2 - 3k) = 0 ⇒$\Rightarrow$ k2 - 3k = 0 ⇒$\Rightarrow$ k (k - 3) = 0 ⇒$\Rightarrow$ k = 0, 3

12. The given equation is 6x2 + x - 2 = 0
    Here, a = 6, b = 1 and, c = - 2
    ∴$\therefore$ D = b2 - 4ac = 1 - 4 ×$\times$ 6 ×$\times$ -2 = 49 > 0
    So, the given equation has real roots, given by
    α=−b+D√2a=−1+49√2×6=−1+712=612=12$\alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-1+\sqrt{49}}{2\times 6}=\frac{-1+7}{12}=\frac{6}{12}=\frac{1}{2}$
    and, β=−b−D√2a=−1−49√2×6=−1−712=−812=−23$\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-1-\sqrt{49}}{2\times 6}=\frac{-1-7}{12}=\frac{-8}{12}=\frac{-2}{3}$

13. The given equation is (x - 3) (2x +1) = x (x+5)
    ⟹$⟹$2x2 + x - 6x - 3 = x2 + 5x
    ⟹$⟹$2x2 - 5x - 3 = x2 + 5x
    ⟹$⟹$x2 - 10x - 3 = 0
    It is in the form of ax2 + bx + c = 0, a≠0$a\ne 0$
    ∴$\therefore$ the given equation is a quadratic equation.

14. Let Shefali's marks in Mathematics = x
    Let Shefali's marks in English = 30 − x
    If, she had got 2 marks more in Mathematics, her marks would be = x + 2
    If, she had got 3 marks less in English, her marks in English would be = 30 – x − 3 = 27 − x
    According to given condition:
    ⇒$\Rightarrow$ (x + 2)(27 − x) = 210
    ⇒27x−x2+54−2x=210$\Rightarrow 27x-x^2+54-2x=210$
    ⇒x2−25x+156=0$\Rightarrow x^2-25x+156=0$
    Comparing quadratic equation x2−25x+156=0$x^2-25x+156=0$ with general form ax2+bx+c=0$a{x}^2+bx+c=0$,
    We get a = 1, b = −25 and c = 156
    Applying Quadratic Formula x=−b±b2−4ac√2a$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
    x=25±(−25)2−4(1)(156)√2×1$x=\frac{25\pm \sqrt{(-25{)}^2-4(1)(156)}}{2\times 1}$
    ⇒25±625−624√2$\Rightarrow \frac{25\pm \sqrt{625-624}}{2}$
    ⇒x=25±1√2$\Rightarrow x=\frac{25\pm \sqrt{1}}{2}$
    ⇒x=25+12,25−12$\Rightarrow x=\frac{25+1}{2},\frac{25-1}{2}$
    ⇒ x = 13, 12
    Therefore, Shefali's marks in Mathematics = 13 or 12
    Shefali's marks in English = 30 – x = 30 – 13 = 17
    Or Shefali's marks in English = 30 – x = 30 – 12 = 18
    Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).

15. Given, 2 is a root of the equation, 3x2+px−8=0$3x^2+px-8=0$
    Putting x=2 in 3x2+px−8=0$x=2\text{ in }3x^2+px-8=0$
    12 + 2p - 8 = 0
    or, p = - 2
    Given, 4x2−2px+k=0$4x^2-2px+k=0$ has equal roots
    4x2+4x+k=0$4x^2+4x+k=0$ has equal roots
    D=b2−4ac=0$D=b^2-4ac=0$
    or, (4)2−4(4)(k)=0$(4{)}^2-4(4)(k)=0$
    or, 16-16k=0
    or, 16k=16
    ∴$\therefore$ k=1

16. Given quadratic equations are;
    x2 + px + q = 0 —(i)
    and, x2 + rx + s = 0 ......(ii)
    Also given ; pr = 2(q + s)........(iii)
    Let D1 and D2 be the discriminant of quadratic equations (i) and (ii) respectively. Then,
    D1 = p2 - 4q and D2 = r2 - 4s
    ⇒$\Rightarrow$ D1+ D2 = p2 - 4q + r2 - 4s = (p2 + r2) - 4(q + s)
    ⇒D1+D2=p2+r2−4(pr2)$\Rightarrow D_1+D_2=p^2+r^2-4\left(\frac{pr}{2}\right)$ $$([from equation (iii)]
    ⇒D1+D2=p2+r2−2pr=(p−r)2≥0$\Rightarrow D_1+D_2=p^2+r^2-2pr=(p-r{)}^2\ge 0$ [∵(p−r)2≥0 for all real p,r]$\left[\because (p-r{)}^2\ge 0\text{ for all real }p,r\right]$
    Now, Since sum of both D2 & D1 is greater than or equal to 0. Hence, both can't be negative.
    ⇒$\Rightarrow$ At least one of D1and D2 is greater than or equal to zero
    Case 1. If D1 ≥ 0, equation (i) has real roots.
    Case 2.If D2 ≥ 0, equation (ii) has real roots.
    Case 3. If D1 & D2 both ≥ 0, then equation (i) & (ii) both have equal roots.
    Clearly, from case 1,2 & 3 at least one given quadratic equations has equal roots. ​​​​

17. Given, speed of boat in still water = 8 Km/hr. Let the speed of the stream be x km/hr. Then,
    Speed of boat in downstream = (8 + x) km/hr
    Speed of boat in upstream = (8 - x) km/hr
    We know that time taken to cover 'd' km with speed 's' km/hr is ds$\frac{d}{s}$

    So,Time taken by the boat to go 15 km upstream =158−x$=\frac{15}{8-x}$ hours.
    &, Time taken by the boat to 22 km downstream =228+x$=\frac{22}{8+x}$ hours.
    It is given that the total time taken by boat to go 15 km upstream & 22 km downstream is 5 hours.
    ∴158−x+228+x=5$\therefore \frac{15}{8-x}+\frac{22}{8+x}=5$
    ⇒15(8+x)+22(8−x)(8−x)(8+x)=5$\Rightarrow \frac{15(8+x)+22(8-x)}{(8-x)(8+x)}=5$
    ⇒120+15x+176−22x82−x2=5$\Rightarrow \frac{120+15x+176-22x}{8^2-x^2}=5$
    ⇒−7x+29664−x2=5$\Rightarrow \frac{-7x+296}{64-x^2}=5$
    ⇒$\Rightarrow$ -7x + 296 = 5(64 - x2)
    ⇒$\Rightarrow$ -7x + 296 = 320 - 5x2
    ⇒$\Rightarrow$ 5x2 - 7x + 296 - 320 = 0
    ⇒$\Rightarrow$ 5x2 - 7x - 24 = 0
    ⇒$\Rightarrow$ 5x2 - 15x + 8x - 24 = 0
    ⇒$\Rightarrow$ 5x(x - 3) + 8(x - 3) = 0
    ⇒$\Rightarrow$ (5x + 8)(x- 3) = 0
    ⇒$\Rightarrow$ x - 3 = 0 [∵$\because$ Speed can not be negative ∴$\therefore$ 5x + 8 ≠$\ne$ 0]
    ⇒$\Rightarrow$ x = 3
    Hence, the speed of the stream is 3 km/hr.

18. Given that a train travelling at a uniform speed for 360 km
    Let the original speed of the train be x km/hr
    Time taken = Distance  Speed =360x$=\frac{\text{ Distance }}{\text{ Speed }}=\frac{360}{x}$
    Time taken at increased speed =360x+5$=\frac{360}{x+5}$ hours.
    According to the question
    360x−360x+5=4860$\frac{360}{x}-\frac{360}{x+5}=\frac{48}{60}$
    360[1x−1x+5]=45$360\left[\frac{1}{x}-\frac{1}{x+5}\right]=\frac{4}{5}$
    or,360(x+5−x)x2+5x=45$or,\frac{360(x+5-x)}{x^2+5x}=\frac{4}{5}$
    or,1800x2+5x=45$or,\frac{1800}{x^2+5x}=\frac{4}{5}$
    ⇒x2+5x−2250=0$\Rightarrow x^2+5x-2250=0$
    ⇒x2+(50−45)x−2250=0$\Rightarrow x^2+(50-45)x-2250=0$
    ⇒x2+50x−45x−2250=0$\Rightarrow x^2+50x-45x-2250=0$
    ⇒(x+50)(x−45)=0$\Rightarrow (x+50)(x-45)=0$
    Either x = - 50 or x = 45
    $$As speed cannot be negative
    ∴$\therefore$ Original speed of train = 45 km/hr.

19. We have the following equation,
    3–√x2+10x+73–√=0$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$
    Now factorise the equation,
    3–√x2+3x+7x+73–√=0$\sqrt{3}{x}^2+3x+7x+7\sqrt{3}=0$
    ⇒3–√x(x+3–√)+7(x+3–√)=0$\Rightarrow \sqrt{3}x(x+\sqrt{3})+7(x+\sqrt{3})=0$
    ⇒(x+3–√)(3–√x+7)=0$\Rightarrow (x+\sqrt{3})(\sqrt{3}x+7)=0$
    ⇒x=−3–√ or x=−73√$\Rightarrow x=-\sqrt{3}\text{ or }x=\frac{-7}{\sqrt{3}}$
    If x=−73√$x=-\frac{7}{\sqrt{3}}$ we need to rationalise it.
    ⇒x=−7×3√3√×3√=−73√3$\Rightarrow x=-\frac{7\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=-\frac{7\sqrt{3}}{3}$
    Therefore, Roots are −3–√,−73√3$-\sqrt{3},-\frac{7\sqrt{3}}{3}$

20. Let x+22x−3$\frac{x+2}{2x-3}$ = y ...(i)
    ∴$\therefore$Given equation becomes,
    2y - 9 ×$\times$ 1y$\frac{1}{y}$ = 3
    ⇒$\Rightarrow$2y2−3y−9=0$2y^2-3y-9=0$
    ⇒$\Rightarrow$ 2y2−6y+3y−9=0$2y^2-6y+3y-9=0$
    ⇒$\Rightarrow$2y(y−3)+3(y−3)=0$2y(y-3)+3(y-3)=0$
    ⇒$\Rightarrow$(2y+3)(y−3)=0$(2y+3)(y-3)=0$
    ⇒$\Rightarrow$y = -32$\frac{3}{2}$ or y = 3
    Putting the value of y in equation (i), we get
    ⇒$\Rightarrow$x+22x−3$\frac{x+2}{2x-3}$ = -32$\frac{3}{2}$ or x+22x−3$\frac{x+2}{2x-3}$ = 3
    ⇒$\Rightarrow$2x+4=−6x+9$2x+4=-6x+9$ or x + 2 = 6x - 9
    ⇒$\Rightarrow$8x = 5 or -5x = -11
    ⇒$\Rightarrow$x = 58$\frac{5}{8}$ or x = 115