Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2.4 Solutions

Question 1. Compare the given quadratic equations to the general form and write values of a, b, c.
(i) x² - 7x + 5 = 0
(ii) 2m² = 5m - 5
(iii) y² = 7y
Answer:
(i) x² - 7x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -7, c = 5

(ii) 2m² = 5m - 5
Therefore, 2m² - 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 2, b = -5, c = 5

(iii) y² = 7y
Therefore, y² - 7y + 0 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = -7, c = 0
In simple words: This question requires identifying coefficients 'a', 'b', and 'c' by rewriting quadratic equations into the standard form \(ax^2 + bx + c = 0\). Once in standard form, directly extract the numerical values for 'a', 'b', and 'c'.

🎯 Exam Tip: Pay close attention to signs when identifying 'b' and 'c'. Ensure all terms are on one side of the equation before comparing.

Question 2. Solve using formula.
(i) x² + 6x + 5 = 0
(ii) x² - 3x - 2 = 0
(iii) 3m² + 2m - 7 = 0
(iv) 5m² - 4m - 2 = 0
(v) y² + \(\frac{1}{3}\)y = 2
(vi) 5x² + 13x + 8 = 0
Answer:
(i) x² + 6x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 6, c = 5
Therefore, b² - 4ac = (6)² - 4 × 1 × 5
= 36 - 20 = 16
x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-6 \pm \sqrt{16}}{2(1)}\)
= \(\frac{-6 \pm 4}{2}\)
= \(\frac{2(-3 \pm 2)}{2}\)
Therefore, x = -3 ± 2
Therefore, x = -3 + 2 or x = -3 - 2
Therefore, x = -1 or x = -5
Therefore, The roots of the given quadratic equation are -1 and -5.

(ii) x² - 3x - 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -3, c = -2
Therefore, b² - 4ac = (-3)² - 4 × 1 × (-2)
= 9 + 8 = 17
x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-(-3) \pm \sqrt{17}}{2(1)}\)
x = \(\frac{3 \pm \sqrt{17}}{2}\)
Therefore, x = \(\frac{3 + \sqrt{17}}{2}\) or x = \(\frac{3 - \sqrt{17}}{2}\)
The roots of the given quadratic equation
are \(\frac{3 + \sqrt{17}}{2}\) and \(\frac{3 - \sqrt{17}}{2}\).

(iii) 3m² + 2m - 7 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 3, b = 2, c = -7
Therefore, b² - 4ac = (2)² - 4 × 3 × (-7)
= 4 + 84 = 88
m = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-2 \pm \sqrt{88}}{2(3)}\)
= \(\frac{-2 \pm \sqrt{4 \times 22}}{6}\)
= \(\frac{-2 \pm 2\sqrt{22}}{6}\)
= \(\frac{2(-1 \pm \sqrt{22})}{6}\)
Therefore, m = \(\frac{-1 \pm \sqrt{22}}{3}\)
Therefore, m = \(\frac{-1 + \sqrt{22}}{3}\) or m = \(\frac{-1 - \sqrt{22}}{3}\)
Therefore, The roots of the given quadratic equation
are \(\frac{-1 + \sqrt{22}}{3}\) and \(\frac{-1 - \sqrt{22}}{3}\).

(iv) 5m² - 4m - 2 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = -4, c = -2
Therefore, b² - 4ac = (-4)² - 4 × 5 × (-2)
= 16 + 40 = 56
m = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-(-4) \pm \sqrt{56}}{2(5)}\)
= \(\frac{4 \pm \sqrt{4 \times 14}}{10}\)
= \(\frac{4 \pm 2\sqrt{14}}{10}\)
= \(\frac{2(2 \pm \sqrt{14})}{10}\)
Therefore, m = \(\frac{2 \pm \sqrt{14}}{5}\)
Therefore, m = \(\frac{2 + \sqrt{14}}{5}\) or m = \(\frac{2 - \sqrt{14}}{5}\)
Therefore, The roots of the given quadratic equation
are \(\frac{2 + \sqrt{14}}{5}\) and \(\frac{2 - \sqrt{14}}{5}\).

(v) y² + \(\frac{1}{3}\)y = 2
Therefore, 3y² + y = 6 ...(Multiplying both sides by 3]
Therefore, 3y² + y - 6 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 3, b = 1, c = -6
Therefore, b² - 4ac = (1)² - 4 × 3 × (-6)
= 1 + 72 = 73
Therefore, y = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-1 \pm \sqrt{73}}{2(3)}\)
y = \(\frac{-1 \pm \sqrt{73}}{6}\)
Therefore, y = \(\frac{-1 + \sqrt{73}}{6}\) or y = \(\frac{-1 - \sqrt{73}}{6}\)
The roots of the given quadratic equation
are \(\frac{-1 + \sqrt{73}}{6}\) and \(\frac{-1 - \sqrt{73}}{6}\).

(vi) 5x² + 13x + 8 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 5, b = 13, c = 8
Therefore, b² - 4ac = (13)² - 4 × 5 × 8
= 169 - 160 = 9
Therefore, b² - 4ac = (13)² - 4 × 5 × 8
= 169 - 160 = 9
x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-13 \pm \sqrt{9}}{2(5)}\)
= \(\frac{-13 \pm 3}{10}\)
Therefore, x = \(\frac{-13 + 3}{10}\) or x = \(\frac{-13 - 3}{10}\)
Therefore, x = \(\frac{-10}{10}\) or x = \(\frac{-16}{10}\)
Therefore, x = -1 or x = \(\frac{-8}{5}\)
The roots of the given quadratic equation are -1 and \(\frac{-8}{5}\).
In simple words: To solve a quadratic equation using the formula, first identify the coefficients a, b, and c. Then, calculate the discriminant \(b^2 - 4ac\). Finally, substitute these values into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the two roots.

🎯 Exam Tip: Remember the quadratic formula. Calculation errors with signs or square roots are common, so double-check each step.

Question 3. With the help of the flow chart given below solve the equation x² + 2√3 x + 3 = 0 using the formula.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ़्लोचार्ट एक द्विघात समीकरण को हल करने के लिए सूत्र विधि के चरणों को दर्शाता है। यह तुलना करने, \(b^2-4ac\) का मान ज्ञात करने, सूत्र लिखने और अंत में 'a', 'b', 'c' के मानों को प्रतिस्थापित करके मूलों को ज्ञात करने के चरणों को क्रमबद्ध करता है।
Answer:
(i) Compare equations
x² + 2√3 x + 3 = 0 and
ax² + bx + c = 0 find
the values of a, b, c.
(ii) Find value of b²-4ac.
(iii) Write formula to solve quadratic equation.
(iv) Substitute values of a, b, c and find roots.

Solution:
(i) x² + 2√3 x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 2√3, c = 3

(ii) b² - 4ac = (2√3)² - 4 × 1 × 3
= 12 - 12
= 0

(iii) x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

(iv) x = \(\frac{-2\sqrt{3} \pm 0}{2(1)}\)
Therefore, x = \(\frac{-2\sqrt{3} + 0}{2}\) or x = \(\frac{-2\sqrt{3} - 0}{2}\)
Therefore, x = -√3 or x = -√3
Therefore, The roots of the given quadratic equation are
-√3 and -√3.
In simple words: This problem guides you through solving a quadratic equation using the formula method by following a specific flowchart. First, compare the given equation to the standard form to find a, b, c, then calculate the discriminant, apply the quadratic formula, and finally substitute the values to find the roots.

🎯 Exam Tip: When \(b^2 - 4ac = 0\), the quadratic equation has two equal real roots. Practice simplifying square roots involving numbers like √88 or √56.

Question 1. Solve the equation 2x² + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. (Textbook pg. no. 43)
Answer:
i. Factorisation method:
2x² + 13x + 15 = 0
Therefore, 2x² + 10x + 3x + 15 = 0
Therefore, 2x(x + 5) + 3(x + 5) = 0
Therefore, (x + 5) (2x + 3) = 0
By using the property, if the product of two numbers is zero, then at least zero, we get
Therefore, x + 5 = 0 or 2x + 3 = 0
Therefore, x = -5 or 2x = -3
Therefore, x = -5 or x = \(\frac{-3}{2}\)
Therefore, The roots of the given quadratic equation are \(\frac{-3}{2}\) and -5.

ii. Completing the square method:
2x² + 13x + 15 = 0
Therefore, x² + \(\frac{13}{2}\)x + \(\frac{15}{2}\) = 0 ... [Dividing both sides by 2]
If x² + \(\frac{13}{2}\)x + k = (x + a)², then
x² + \(\frac{13}{2}\)x + k = x² + 2ax + a²
Comparing the coefficients, we get
\(\frac{13}{2}\) = 2a and k = a²
Therefore, a = \(\frac{13}{4}\) and k = (\(\frac{13}{4}\))² = \(\frac{169}{16}\)
Now, x² + \(\frac{13}{2}\)x + \(\frac{15}{2}\) = 0
Therefore, x² + \(\frac{13}{2}\)x + \(\frac{169}{16}\) - \(\frac{169}{16}\) + \(\frac{15}{2}\) = 0
Therefore, (x + \(\frac{13}{4}\))² + (\(\frac{-169 + 120}{16}\)) = 0
Therefore, (x + \(\frac{13}{4}\))² - \(\frac{49}{16}\) = 0
Therefore, (x + \(\frac{13}{4}\))² = \(\frac{49}{16}\)
Taking square root of both sides, we get
x + \(\frac{13}{4}\) = ±\(\frac{7}{4}\)
Therefore, x + \(\frac{13}{4}\) = \(\frac{7}{4}\) or x + \(\frac{13}{4}\) = \(\frac{-7}{4}\)
Therefore, x = \(\frac{7}{4}\) - \(\frac{13}{4}\) or x = \(\frac{-7}{4}\) - \(\frac{13}{4}\)
Therefore, x = \(\frac{7 - 13}{4}\) or x = \(\frac{-7 - 13}{4}\)
Therefore, x = \(\frac{-6}{4}\) or x = \(\frac{-20}{4}\)
Therefore, x = \(\frac{-3}{2}\) or x = -5
Therefore, The roots of the given quadratic equation are \(\frac{-3}{2}\) and -5.

iii. Formula method:
2x² + 13x + 15 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 2, b = 13, c = 15
Therefore, b² - 4ac = (13)² - 4 × 2 × 15
= 169 - 120 = 49
x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
= \(\frac{-13 \pm \sqrt{49}}{2(2)}\)
= \(\frac{-13 \pm 7}{4}\)
Therefore, x = \(\frac{-13 + 7}{4}\) or x = \(\frac{-13 - 7}{4}\)
Therefore, x = \(\frac{-6}{4}\) or x = \(\frac{-20}{4}\)
Therefore, x = \(\frac{-3}{2}\) or x = -5
Therefore, The roots of the given quadratic equation are \(\frac{-3}{2}\) and -5.
Therefore, By all the above three methods, we get the same roots of the given quadratic equation.
In simple words: This question demonstrates three different methods - factorization, completing the square, and the quadratic formula - to solve the same quadratic equation. The objective is to show that regardless of the method used, a quadratic equation will yield the same set of roots.

🎯 Exam Tip: Mastery of all three methods is crucial. The factorization method is quick for simple equations, while the quadratic formula is universal. Completing the square helps in understanding the derivation of the formula.