Class 10 Chapter 4 Quadratic Equations

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Quadratic Equations

Chapter 4 of Class 10 Mathematics typically deals with Quadratic Equations. Let’s work through all the exercises and then create some additional questions with solutions.

Chapter 4: Quadratic Equations

A quadratic equation is an equation of the form:ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0

Where:

  • aaa, bbb, and ccc are constants and a≠0a \neq 0a=0
  • xxx is the variable

Exercise 4.1

Solve the following quadratic equations:

  1. x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0Solution: Factorizing the quadratic equation:x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0 (x−2)(x−3)=0(x – 2)(x – 3) = 0(x−2)(x−3)=0So, x=2x = 2x=2 or x=3x = 3x=3.Answer: x=2x = 2x=2 or x=3x = 3x=3.
  1. 2×2−3x−5=02x^2 – 3x – 5 = 02×2−3x−5=0

Solution: We can solve this using the quadratic formula: The quadratic formula is:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​For 2×2−3x−5=02x^2 – 3x – 5 = 02×2−3x−5=0, we have a=2a = 2a=2, b=−3b = -3b=−3, and c=−5c = -5c=−5.Using the quadratic formula:x=−(−3)±(−3)2−4(2)(−5)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-5)}}{2(2)}x=2(2)−(−3)±(−3)2−4(2)(−5)​​ x=3±9+404x = \frac{3 \pm \sqrt{9 + 40}}{4}x=43±9+40​​ x=3±494x = \frac{3 \pm \sqrt{49}}{4}x=43±49​​ x=3±74x = \frac{3 \pm 7}{4}x=43±7​So, the two possible solutions are:x=3+74=104=2.5orx=3−74=−44=−1x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5 \quad \text{or} \quad x = \frac{3 – 7}{4} = \frac{-4}{4} = -1x=43+7​=410​=2.5orx=43−7​=4−4​=−1Answer: x=2.5x = 2.5x=2.5 or x=−1x = -1x=−1.

  1. x2+4x−21=0x^2 + 4x – 21 = 0x2+4x−21=0

Solution: Using the quadratic formula:x=−4±42−4(1)(−21)2(1)x = \frac{-4 \pm \sqrt{4^2 – 4(1)(-21)}}{2(1)}x=2(1)−4±42−4(1)(−21)​​ x=−4±16+842x = \frac{-4 \pm \sqrt{16 + 84}}{2}x=2−4±16+84​​ x=−4±1002x = \frac{-4 \pm \sqrt{100}}{2}x=2−4±100​​ x=−4±102x = \frac{-4 \pm 10}{2}x=2−4±10​So, the two possible solutions are:x=−4+102=62=3orx=−4−102=−142=−7x = \frac{-4 + 10}{2} = \frac{6}{2} = 3 \quad \text{or} \quad x = \frac{-4 – 10}{2} = \frac{-14}{2} = -7x=2−4+10​=26​=3orx=2−4−10​=2−14​=−7Answer: x=3x = 3x=3 or x=−7x = -7x=−7.

Exercise 4.2

Solve the following quadratic equations by completing the square:

  1. x2+6x−7=0x^2 + 6x – 7 = 0x2+6x−7=0

Solution: Step 1: Move the constant term to the other side:x2+6x=7x^2 + 6x = 7×2+6x=7Step 2: Complete the square. Take half of 6, square it, and add it to both sides:(62)2=9\left(\frac{6}{2}\right)^2 = 9(26​)2=9Now, add 9 to both sides:x2+6x+9=7+9x^2 + 6x + 9 = 7 + 9×2+6x+9=7+9 (x+3)2=16(x + 3)^2 = 16(x+3)2=16Step 3: Solve for xxx:x+3=±4x + 3 = \pm 4x+3=±4So, x=−3+4=1x = -3 + 4 = 1x=−3+4=1 or x=−3−4=−7x = -3 – 4 = -7x=−3−4=−7.Answer: x=1x = 1x=1 or x=−7x = -7x=−7.

  1. x2−8x+10=0x^2 – 8x + 10 = 0x2−8x+10=0

Solution: Step 1: Move the constant term to the other side:x2−8x=−10x^2 – 8x = -10×2−8x=−10Step 2: Complete the square. Take half of -8, square it, and add it to both sides:(−82)2=16\left(\frac{-8}{2}\right)^2 = 16(2−8​)2=16Add 16 to both sides:x2−8x+16=−10+16x^2 – 8x + 16 = -10 + 16×2−8x+16=−10+16 (x−4)2=6(x – 4)^2 = 6(x−4)2=6Step 3: Solve for xxx:x−4=±6x – 4 = \pm \sqrt{6}x−4=±6​So, x=4+6x = 4 + \sqrt{6}x=4+6​ or x=4−6x = 4 – \sqrt{6}x=4−6​.Answer: x=4+6x = 4 + \sqrt{6}x=4+6​ or x=4−6x = 4 – \sqrt{6}x=4−6​.

Exercise 4.3

Word problems on quadratic equations:

  1. A rectangle’s length is 2 meters more than its width. If the area of the rectangle is 120 square meters, find its length and width.

Solution: Let the width of the rectangle be xxx meters. Then, the length is x+2x + 2x+2 meters.The area of the rectangle is given by:Area=Length×Width\text{Area} = \text{Length} \times \text{Width}Area=Length×Width 120=(x+2)×x120 = (x + 2) \times x120=(x+2)×xExpanding:120=x2+2×120 = x^2 + 2×120=x2+2xRearranging the equation:x2+2x−120=0x^2 + 2x – 120 = 0x2+2x−120=0Solve using the quadratic formula:x=−2±22−4(1)(−120)2(1)x = \frac{-2 \pm \sqrt{2^2 – 4(1)(-120)}}{2(1)}x=2(1)−2±22−4(1)(−120)​​ x=−2±4+4802x = \frac{-2 \pm \sqrt{4 + 480}}{2}x=2−2±4+480​​ x=−2±4842x = \frac{-2 \pm \sqrt{484}}{2}x=2−2±484​​ x=−2±222x = \frac{-2 \pm 22}{2}x=2−2±22​So, x=−2+222=10x = \frac{-2 + 22}{2} = 10x=2−2+22​=10 or x=−2−222=−12x = \frac{-2 – 22}{2} = -12x=2−2−22​=−12.Since the width cannot be negative, we take x=10x = 10x=10.Therefore, the width is 10 meters, and the length is 10+2=1210 + 2 = 1210+2=12 meters.Answer: Width = 10 meters, Length = 12 meters.

Additional Questions

  1. Solve the quadratic equation: x2+5x−6=0x^2 + 5x – 6 = 0x2+5x−6=0Solution: Factorizing the equation:x2+5x−6=0x^2 + 5x – 6 = 0x2+5x−6=0 (x+6)(x−1)=0(x + 6)(x – 1) = 0(x+6)(x−1)=0So, x=−6x = -6x=−6 or x=1x = 1x=1.Answer: x=−6x = -6x=−6 or x=1x = 1x=1.
  1. Solve the quadratic equation: 2×2+7x−3=02x^2 + 7x – 3 = 02×2+7x−3=0Solution: Using the quadratic formula:x=−7±72−4(2)(−3)2(2)x = \frac{-7 \pm \sqrt{7^2 – 4(2)(-3)}}{2(2)}x=2(2)−7±72−4(2)(−3)​​ x=−7±49+244x = \frac{-7 \pm \sqrt{49 + 24}}{4}x=4−7±49+24​​ x=−7±734x = \frac{-7 \pm \sqrt{73}}{4}x=4−7±73​​So, the two solutions are:x=−7+734orx=−7−734x = \frac{-7 + \sqrt{73}}{4} \quad \text{or} \quad x = \frac{-7 – \sqrt{73}}{4}x=4−7+73​​orx=4−7−73​​Answer: x=−7+734x = \frac{-7 + \sqrt{73}}{4}x=4−7+73​​ or x=−7−734x = \frac{-7 – \sqrt{73}}{4}x=4−7−73​​.

1. Solve the equation:

x2−8x+16=0x^2 – 8x + 16 = 0x2−8x+16=0

Solution: This is a perfect square trinomial. It can be factored as:(x−4)2=0(x – 4)^2 = 0(x−4)2=0

Taking the square root of both sides:x−4=0x – 4 = 0x−4=0 x=4x = 4x=4

Answer: x=4x = 4x=4.

2. Solve the equation:

3×2−7x+2=03x^2 – 7x + 2 = 03×2−7x+2=0

Solution: We use the quadratic formula for this:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​

Here, a=3a = 3a=3, b=−7b = -7b=−7, and c=2c = 2c=2.

Substitute into the formula:x=−(−7)±(−7)2−4(3)(2)2(3)x = \frac{-(-7) \pm \sqrt{(-7)^2 – 4(3)(2)}}{2(3)}x=2(3)−(−7)±(−7)2−4(3)(2)​​ x=7±49−246x = \frac{7 \pm \sqrt{49 – 24}}{6}x=67±49−24​​ x=7±256x = \frac{7 \pm \sqrt{25}}{6}x=67±25​​ x=7±56x = \frac{7 \pm 5}{6}x=67±5​

So, the two solutions are:x=7+56=126=2orx=7−56=26=13x = \frac{7 + 5}{6} = \frac{12}{6} = 2 \quad \text{or} \quad x = \frac{7 – 5}{6} = \frac{2}{6} = \frac{1}{3}x=67+5​=612​=2orx=67−5​=62​=31​

Answer: x=2x = 2x=2 or x=13x = \frac{1}{3}x=31​.

3. Solve the equation by factorization:

x2+5x−24=0x^2 + 5x – 24 = 0x2+5x−24=0

Solution: We need to factor the quadratic. We want two numbers that multiply to −24-24−24 and add up to 555. These numbers are 888 and −3-3−3, because:8×(−3)=−24and8+(−3)=58 \times (-3) = -24 \quad \text{and} \quad 8 + (-3) = 58×(−3)=−24and8+(−3)=5

So, we can factor the equation as:(x+8)(x−3)=0(x + 8)(x – 3) = 0(x+8)(x−3)=0

Now solve for xxx:x+8=0orx−3=0x + 8 = 0 \quad \text{or} \quad x – 3 = 0x+8=0orx−3=0 x=−8orx=3x = -8 \quad \text{or} \quad x = 3x=−8orx=3

Answer: x=−8x = -8x=−8 or x=3x = 3x=3.

4. Solve the equation using the quadratic formula:

4×2−12x+9=04x^2 – 12x + 9 = 04×2−12x+9=0

Solution: We use the quadratic formula here. For 4×2−12x+9=04x^2 – 12x + 9 = 04×2−12x+9=0, a=4a = 4a=4, b=−12b = -12b=−12, and c=9c = 9c=9.

Substitute into the quadratic formula:x=−(−12)±(−12)2−4(4)(9)2(4)x = \frac{-(-12) \pm \sqrt{(-12)^2 – 4(4)(9)}}{2(4)}x=2(4)−(−12)±(−12)2−4(4)(9)​​ x=12±144−1448x = \frac{12 \pm \sqrt{144 – 144}}{8}x=812±144−144​​ x=12±08x = \frac{12 \pm \sqrt{0}}{8}x=812±0​​ x=12±08x = \frac{12 \pm 0}{8}x=812±0​ x=128=32x = \frac{12}{8} = \frac{3}{2}x=812​=23​

Answer: x=32x = \frac{3}{2}x=23​.

5. Solve the equation:

x2−6x−72=0x^2 – 6x – 72 = 0x2−6x−72=0

Solution: We will factor this quadratic. We need two numbers that multiply to −72-72−72 and add up to −6-6−6. These numbers are −12-12−12 and 666, because:−12×6=−72and−12+6=−6-12 \times 6 = -72 \quad \text{and} \quad -12 + 6 = -6−12×6=−72and−12+6=−6

So, we factor the equation as:(x−12)(x+6)=0(x – 12)(x + 6) = 0(x−12)(x+6)=0

Now solve for xxx:x−12=0orx+6=0x – 12 = 0 \quad \text{or} \quad x + 6 = 0x−12=0orx+6=0 x=12orx=−6x = 12 \quad \text{or} \quad x = -6x=12orx=−6

Answer: x=12x = 12x=12 or x=−6x = -6x=−6.

6. Solve the quadratic equation:

2×2+5x−3=02x^2 + 5x – 3 = 02×2+5x−3=0

Solution: We will use the quadratic formula here, where a=2a = 2a=2, b=5b = 5b=5, and c=−3c = -3c=−3.

Substitute into the quadratic formula:x=−5±52−4(2)(−3)2(2)x = \frac{-5 \pm \sqrt{5^2 – 4(2)(-3)}}{2(2)}x=2(2)−5±52−4(2)(−3)​​ x=−5±25+244x = \frac{-5 \pm \sqrt{25 + 24}}{4}x=4−5±25+24​​ x=−5±494x = \frac{-5 \pm \sqrt{49}}{4}x=4−5±49​​ x=−5±74x = \frac{-5 \pm 7}{4}x=4−5±7​

So, the two solutions are:x=−5+74=24=12orx=−5−74=−124=−3x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-5 – 7}{4} = \frac{-12}{4} = -3x=4−5+7​=42​=21​orx=4−5−7​=4−12​=−3

Answer: x=12x = \frac{1}{2}x=21​ or x=−3x = -3x=−3.

7. Solve the equation:

x2+7x−12=0x^2 + 7x – 12 = 0x2+7x−12=0

Solution: We will factor this quadratic. We need two numbers that multiply to −12-12−12 and add up to 777. These numbers are 121212 and −1-1−1, because:12×(−1)=−12and12+(−1)=712 \times (-1) = -12 \quad \text{and} \quad 12 + (-1) = 712×(−1)=−12and12+(−1)=7

So, we factor the equation as:(x+12)(x−1)=0(x + 12)(x – 1) = 0(x+12)(x−1)=0

Now solve for xxx:x+12=0orx−1=0x + 12 = 0 \quad \text{or} \quad x – 1 = 0x+12=0orx−1=0 x=−12orx=1x = -12 \quad \text{or} \quad x = 1x=−12orx=1

Answer: x=−12x = -12x=−12 or x=1x = 1x=1.

8. Find the roots of the equation:

5×2−15x=05x^2 – 15x = 05×2−15x=0

Solution: We can factor the equation:5x(x−3)=05x(x – 3) = 05x(x−3)=0

Now, set each factor equal to zero:5x=0orx−3=05x = 0 \quad \text{or} \quad x – 3 = 05x=0orx−3=0x=0orx=3x = 0 \quad \text{or} \quad x = 3x=0orx=3

Answer: x=0x = 0x=0 or x=3x = 3x=3.

9. Solve the quadratic equation:

x2+4x−5=0x^2 + 4x – 5 = 0x2+4x−5=0

Solution: We need to factor the quadratic. We need two numbers that multiply to −5-5−5 and add up to 444. These numbers are 555 and −1-1−1, because:5×(−1)=−5and5+(−1)=45 \times (-1) = -5 \quad \text{and} \quad 5 + (-1) = 45×(−1)=−5and5+(−1)=4

So, we can factor the equation as:(x+5)(x−1)=0(x + 5)(x – 1) = 0(x+5)(x−1)=0

Now solve for xxx:x+5=0orx−1=0x + 5 = 0 \quad \text{or} \quad x – 1 = 0x+5=0orx−1=0x=−5orx=1x = -5 \quad \text{or} \quad x = 1x=−5orx=1

Answer: x=−5x = -5x=−5 or x=1x = 1x=1.

10. Solve for xxx:

2×2+9x−5=02x^2 + 9x – 5 = 02×2+9x−5=0

Solution: We use the quadratic formula here, where a=2a = 2a=2, b=9b = 9b=9, and c=−5c = -5c=−5.

Substitute into the quadratic formula:x=−9±92−4(2)(−5)2(2)x = \frac{-9 \pm \sqrt{9^2 – 4(2)(-5)}}{2(2)}x=2(2)−9±92−4(2)(−5)​​x=−9±81+404x = \frac{-9 \pm \sqrt{81 + 40}}{4}x=4−9±81+40​​x=−9±1214x = \frac{-9 \pm \sqrt{121}}{4}x=4−9±121​​x=−9±114x = \frac{-9 \pm 11}{4}x=4−9±11​

So, the two solutions are:x=−9+114=24=12orx=−9−114=−204=−5x = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-9 – 11}{4} = \frac{-20}{4} = -5x=4−9+11​=42​=21​orx=4−9−11​=4−20​=−5

Answer: x=12x = \frac{1}{2}x=21​ or x=−5x = -5x=−5.

11. Find the roots of the quadratic equation:

x2−10x+25=0x^2 – 10x + 25 = 0x2−10x+25=0

Solution: This is a perfect square trinomial:(x−5)2=0(x – 5)^2 = 0(x−5)2=0

Taking the square root of both sides:x−5=0x – 5 = 0x−5=0x=5x = 5x=5

Answer: x=5x = 5x=5.

12. Solve the equation:

3×2+2x−1=03x^2 + 2x – 1 = 03×2+2x−1=0

Solution: We will use the quadratic formula. Here, a=3a = 3a=3, b=2b = 2b=2, and c=−1c = -1c=−1.

Substitute into the quadratic formula:x=−2±22−4(3)(−1)2(3)x = \frac{-2 \pm \sqrt{2^2 – 4(3)(-1)}}{2(3)}x=2(3)−2±22−4(3)(−1)​​x=−2±4+126x = \frac{-2 \pm \sqrt{4 + 12}}{6}x=6−2±4+12​​x=−2±166x = \frac{-2 \pm \sqrt{16}}{6}x=6−2±16​​x=−2±46x = \frac{-2 \pm 4}{6}x=6−2±4​

So, the two solutions are:x=−2+46=26=13orx=−2−46=−66=−1x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-2 – 4}{6} = \frac{-6}{6} = -1x=6−2+4​=62​=31​orx=6−2−4​=6−6​=−1

Answer: x=13x = \frac{1}{3}x=31​ or x=−1x = -1x=−1.

13. Solve the equation:

x2+3x−10=0x^2 + 3x – 10 = 0x2+3x−10=0

Solution: We will factor the quadratic. We need two numbers that multiply to −10-10−10 and add up to 333. These numbers are 555 and −2-2−2, because:5×(−2)=−10and5+(−2)=35 \times (-2) = -10 \quad \text{and} \quad 5 + (-2) = 35×(−2)=−10and5+(−2)=3

So, we can factor the equation as:(x+5)(x−2)=0(x + 5)(x – 2) = 0(x+5)(x−2)=0

Now solve for xxx:x+5=0orx−2=0x + 5 = 0 \quad \text{or} \quad x – 2 = 0x+5=0orx−2=0x=−5orx=2x = -5 \quad \text{or} \quad x = 2x=−5orx=2

Answer: x=−5x = -5x=−5 or x=2x = 2x=2.

14. Solve for xxx:

x2+14x+49=0x^2 + 14x + 49 = 0x2+14x+49=0

Solution: This is a perfect square trinomial:(x+7)2=0(x + 7)^2 = 0(x+7)2=0

Taking the square root of both sides:x+7=0x + 7 = 0x+7=0x=−7x = -7x=−7

Answer: x=−7x = -7x=−7.

15. Solve the quadratic equation:

x2+6x+9=0x^2 + 6x + 9 = 0x2+6x+9=0

Solution: This is a perfect square trinomial:(x+3)2=0(x + 3)^2 = 0(x+3)2=0

Taking the square root of both sides:x+3=0x + 3 = 0x+3=0x=−3x = -3x=−3

Answer: x=−3x = -3x=−3.

16. Solve the equation:

2×2+3x−2=02x^2 + 3x – 2 = 02×2+3x−2=0

Solution: We will use the quadratic formula. Here, a=2a = 2a=2, b=3b = 3b=3, and c=−2c = -2c=−2.

Substitute into the quadratic formula:x=−3±32−4(2)(−2)2(2)x = \frac{-3 \pm \sqrt{3^2 – 4(2)(-2)}}{2(2)}x=2(2)−3±32−4(2)(−2)​​x=−3±9+164x = \frac{-3 \pm \sqrt{9 + 16}}{4}x=4−3±9+16​​x=−3±254x = \frac{-3 \pm \sqrt{25}}{4}x=4−3±25​​x=−3±54x = \frac{-3 \pm 5}{4}x=4−3±5​

So, the two solutions are:x=−3+54=24=12orx=−3−54=−84=−2x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-3 – 5}{4} = \frac{-8}{4} = -2x=4−3+5​=42​=21​orx=4−3−5​=4−8​=−2

Answer: x=12x = \frac{1}{2}x=21​ or x=−2x = -2x=−2.

17. Solve the quadratic equation:

x2+8x+15=0x^2 + 8x + 15 = 0x2+8x+15=0

Solution: We need to factor the quadratic. We need two numbers that multiply to 151515 and add up to 888. These numbers are 333 and 555, because:3×5=15and3+5=83 \times 5 = 15 \quad \text{and} \quad 3 + 5 = 83×5=15and3+5=8

So, we can factor the equation as:(x+3)(x+5)=0(x + 3)(x + 5) = 0(x+3)(x+5)=0

Now solve for xxx:x+3=0orx+5=0x + 3 = 0 \quad \text{or} \quad x + 5 = 0x+3=0orx+5=0x=−3orx=−5x = -3 \quad \text{or} \quad x = -5x=−3orx=−5

Answer: x=−3x = -3x=−3 or x=−5x = -5x=−5.

18. Solve for xxx:

x2−9x−18=0x^2 – 9x – 18 = 0x2−9x−18=0

Solution: We will factor the quadratic. We need two numbers that multiply to −18-18−18 and add up to −9-9−9. These numbers are −12-12−12 and 333, because:−12×3=−18and−12+3=−9-12 \times 3 = -18 \quad \text{and} \quad -12 + 3 = -9−12×3=−18and−12+3=−9

So, we can factor the equation as:(x−12)(x+3)=0(x – 12)(x + 3) = 0(x−12)(x+3)=0

Now solve for xxx:x−12=0orx+3=0x – 12 = 0 \quad \text{or} \quad x + 3 = 0x−12=0orx+3=0x=12orx=−3x = 12 \quad \text{or} \quad x = -3x=12orx=−3

Answer: x=12x = 12x=12 or x=−3x = -3x=−3.

19. Solve the equation using the quadratic formula:

x2+2x−3=0x^2 + 2x – 3 = 0x2+2x−3=0

Solution: We will use the quadratic formula. Here, a=1a = 1a=1, b=2b = 2b=2, and c=−3c = -3c=−3.

Substitute into the quadratic formula:x=−2±22−4(1)(−3)2(1)x = \frac{-2 \pm \sqrt{2^2 – 4(1)(-3)}}{2(1)}x=2(1)−2±22−4(1)(−3)​​x=−2±4+122x = \frac{-2 \pm \sqrt{4 + 12}}{2}x=2−2±4+12​​x=−2±162x = \frac{-2 \pm \sqrt{16}}{2}x=2−2±16​​x=−2±42x = \frac{-2 \pm 4}{2}x=2−2±4​

So, the two solutions are:x=−2+42=22=1orx=−2−42=−62=−3x = \frac{-2 + 4}{2} = \frac{2}{2} = 1 \quad \text{or} \quad x = \frac{-2 – 4}{2} = \frac{-6}{2} = -3x=2−2+4​=22​=1orx=2−2−4​=2−6​=−3

Answer: x=1x = 1x=1 or x=−3x = -3x=−3.

20. Solve for xxx:

x2−4x−21=0x^2 – 4x – 21 = 0x2−4x−21=0

Solution: We need to factor the quadratic. We need two numbers that multiply to −21-21−21 and add up to −4-4−4. These numbers are −7-7−7 and 333, because:−7×3=−21and−7+3=−4-7 \times 3 = -21 \quad \text{and} \quad -7 + 3 = -4−7×3=−21and−7+3=−4

So, we can factor the equation as:(x−7)(x+3)=0(x – 7)(x + 3) = 0(x−7)(x+3)=0

Now solve for xxx:x−7=0orx+3=0x – 7 = 0 \quad \text{or} \quad x + 3 = 0x−7=0orx+3=0x=7orx=−3x = 7 \quad \text{or} \quad x = -3x=7orx=−3

Answer: x=7x = 7x=7 or x=−3x = -3x=−3.

These are the solutions for all 20 questions! Let me know if you need any further clarification on any step.

Also Read: Class 10 Chapter 3 Maths Pair of Linear Equations in Two Variables

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