Exercise 1.2 Class 10 for Confident Problem Solving

Q1. part i ,part ii , part iii , part iv , part v , part vi , part vii , part viii , part ix , part x

In Exercise 1.2 Class 10, students will master the art of solving quadratic equations using the quadratic formula. This powerful formula helps simplify even the most complex quadratic equations, providing a straightforward method to find their roots. By practicing the steps involved, you’ll gain confidence in solving equations efficiently, making this exercise a key step towards mastering quadratic equations and strengthening your algebra skills. To make your learning more efficient the direct links to the individual parts of Question 1 are given above. This way, you can skip the hassle of scrolling through the entire exercise 1.2 class 10 and quickly jump to the specific part you’re looking for!

Question 1 of Exercise 1.2 Class 10: Solve following equations using quadratic formula

(i) 2 − x² = 7x

$$\begin{array}{l}2–x^2=7x\\ –x^2–7x+2=0\\ x^2+7x–2=0\\ x=\frac{–b\pm \sqrt{b^2–4ac}}{2a}\\ x=\frac{–7\pm \sqrt{49+8}}{2}\\ x=\frac{–7\pm \sqrt{57}}{2}\\ x=\frac{–7+\sqrt{57}}{2},x=\frac{–7–\sqrt{57}}{2}\end{array}$$

(ii) 5x² + 8x + 1 = 0

$$\begin{array}{l}\mathrm{Using Quadratic Formula},x=\frac{–b\pm \sqrt{b^2–4ac}}{2a}\\ \mathrm{Here},(a=5),(b=8),(c=1):\\ x=\frac{–8\pm \sqrt{64–20}}{10}=\frac{–8\pm \sqrt{44}}{10}\\ \frac{–8\pm 2\sqrt{11}}{10}\\ \mathrm{Solution Set}=\frac{–4\pm \sqrt{11}}{5}\end{array}$$

(iii)√3x² + x = 4√3

$$\begin{array}{l}\sqrt{3}{x}^2+x–4\sqrt{3}=0\\ \mathrm{Here},(a=\sqrt{3}),(b=1),(c=–4\sqrt{3}):\\ x=\frac{–1\pm \sqrt{1+48}}{2\sqrt{3}}=\frac{–1\pm \sqrt{49}}{2\sqrt{3}}=\frac{–1\pm 7}{2\sqrt{3}}\\ x=\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3} \mathrm{or} x=\frac{–}{8}\\ \mathrm{Solution Set of x}=\{\sqrt{3},\frac{–8}{2\sqrt{3}}\}\end{array}$$

(iv) 4x² – 14 = 3x

$$\begin{array}{l}4x^2–3x–14=0\\ \mathrm{Here},(a=4),(b=–3),(c=–14):\\ x=\frac{3\pm \sqrt{9+224}}{8}\\ \mathrm{Solution Set of x}=\frac{3\pm \sqrt{233}}{8}\end{array}$$

(v) 6x² – 3 – 7x = 0

$$\begin{array}{l}6x^2–7x–3=0\\ \mathrm{Here},(a=6),(b=–7),(c=–3)\\ x=\frac{7\pm \sqrt{49+72}}{12}\\ x=\frac{7\pm \sqrt{121}}{12}\\ x=\frac{18}{12}=\frac{3}{2} \mathrm{or} x=\frac{–}{4}=–\frac{1}{3}\\ \mathrm{Solution Set of x}=\frac{–}{4}, –\frac{1}{3}\end{array}$$

(vi) 3x² + 8x + 2 = 0

(vii)3 / (x – 6) – 4 / (x – 5) = 1

$\frac{3}{x–6}–\frac{4}{x–5}=1$

$3(x–5)–4(x–6)=(x–6)(x–5)$

$3x–15–4x+24=x^2–11x+30$

$x^2–10x+27=0$

$x=\frac{–(–10)\pm \sqrt{(–102^{}–4(1\left)\right(27)}}{2\left(1\right)}$

$x=\frac{10\pm \sqrt{16}}{2}$

$x=\frac{10\pm 4}{2}$

$x=7orx=3$

(viii) \( \frac{x+2}{x-1} – \frac{4-x}{2x} = 2\frac{1}{3} \)

$\frac{2x(x+2)–(4–x)(x–1)}{2x(x–1)}=\frac{7}{3}$

$2x^2+4x–4x+4+x^2–x=\frac{14}{3}{x}^2–\frac{14}{3}x$

$3x^2–11x+4=\frac{14}{3}{x}^2–\frac{14}{3}x$

$9x^2–33x+12=14x^2–14x$

$5x^2–19x+12=0$

$x=\frac{–(–19)\pm \sqrt{(–192^{}–4(5\left)\right(12)}}{2\left(5\right)}$

$x=\frac{19\pm \sqrt{361–240}}{10}$

$x=\frac{19\pm \sqrt{121}}{10}$

$x=\frac{19\pm 11}{10}$

$x=\frac{30}{10}orx=\frac{8}{10}$

$x=3orx=\frac{4}{5}$

(ix) \(\frac{a}{x-b} + \frac{b}{x-a} = 2\)

\(\frac{a}{x-b} + \frac{b}{x-a} = 2\)

\(a(x-a) + b(x-b) = 2(x-b)(x-a)\)

\(ax – a^2 + bx – b^2 = 2x^2 – 2(a+b)x + 2ab\)

\(2x^2 – 3(a+b)x + (a^2 + b^2 + 2ab) = 0\)

\(2x^2 – 3(a+b)x + (a+b)^2 = 0\)

\(x = \frac{-3(a+b) \pm \sqrt{(-3(a+b))^2 – 4(2)(a+b)^2}}{2(2)}\)

\(x = \frac{3(a+b) \pm \sqrt{9(a+b)^2 – 8(a+b)^2}}{4}\)

\(x = \frac{3(a+b) \pm (a+b)}{4}\)

\(x = a + b \text{ or } x = \frac{a+b}{2}\)

(X) \( -(l+m) – lx^2 + (2l+m)x = 0 \)

To solve the quadratic equation \[ -(l + m) – lx^2 + (2l + m) x = 0 \] rewrite it in the standard form \( ax^2 + bx + c = 0 \): \[ -lx^2 + (2l + m)x – (l + m) = 0\] \[a = -l,b = 2l + m,c = -(l + m)\] \[x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\] \[x = \frac{-(2l + m) \pm \sqrt{(2l + m)^2 – 4(-l)(-(l + m))}}{2(-l)}\] \[x = \frac{-(2l + m) \pm \sqrt{(2l + m)^2 – 4l(l + m)}}{-2l}\] \[x = \frac{-(2l + m) \pm \sqrt{4l^2 + 4lm + m^2 – 4l^2 – 4lm}}{-2l}\] \[x = \frac{-(2l + m) \pm \sqrt{m^2}}{-2l}= \frac{-(2l + m) \pm m}{-2l}\] Separate into two solutions: \[x = \frac{-(2l + m) + m}{-2l} = \frac{-2l}{-2l} = 1\] \[x = \frac{-(2l + m) – m}{-2l} = \frac{-2l – 2m}{-2l} = 1 + \frac{m}{l}\] The solution set is: \[{ 1, \; 1 + \frac{m}{l} }\]

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Derivation of Quadratic Formula

Continue your learning journey with the solutions for Exercise 1.3 and deepen your understanding of quadratic equations.

Here’s a complete video lecture of Exercise 1.2 Class 10.

We hope that with the solutions provided in Exercise 1.2 Class 10 and the accompanying video lecture, you have gained a solid understanding of solving quadratic equations using the quadratic formula. Purple math has a calculator that solves quadratic equation for you have a look at it. Keep practicing, and you’ll continue to build your confidence and skills for tackling more challenging problems in the future!