Quadratic Equations — NCERT Solutions, AI Tutor & Practice

Expand the left side: $(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$.

Expand the right side: $x(x+5) = x^2 + 5x$.

Bring all terms to one side: $2x^2 - 5x - 3 - x^2 - 5x = 0$ $x^2 - 10x - 3 = 0$

The highest power of $x$ is $2$ and the coefficient of $x^2$ is $1 \neq 0$. Hence it is a quadratic equation in standard form $x^2 - 10x - 3 = 0$.

We split the middle term. We need two numbers whose product is $6 \times (-2) = -12$ and whose sum is $-1$. These numbers are $-4$ and $3$.

$6x^2 - 4x + 3x - 2 = 0$ $2x(3x - 2) + 1(3x - 2) = 0$ $(3x - 2)(2x + 1) = 0$

Setting each factor to zero: $3x - 2 = 0 \Rightarrow x = \tfrac{2}{3}$ $2x + 1 = 0 \Rightarrow x = -\tfrac{1}{2}$

Thus the roots are $x = \dfrac{2}{3}$ and $x = -\dfrac{1}{2}$.

Here $a = 2$, $b = -7$, $c = 3$.

Discriminant: $b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25$.

Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$

So $x = \dfrac{7+5}{4} = 3$ or $x = \dfrac{7-5}{4} = \dfrac{1}{2}$.

Roots: $x = 3$ and $x = \dfrac{1}{2}$.

Start with $x^2 + 4x - 5 = 0$, so $x^2 + 4x = 5$.

Half of the coefficient of $x$ is $\frac{4}{2} = 2$, and $2^2 = 4$. Add $4$ to both sides: $x^2 + 4x + 4 = 5 + 4$ $(x + 2)^2 = 9$

Taking square roots: $x + 2 = \pm 3$ $x = -2 + 3 = 1 \quad \text{or} \quad x = -2 - 3 = -5$

Roots: $x = 1$ and $x = -5$.

Divide throughout by $3$: $x^2 - \frac{5}{3}x + \frac{2}{3} = 0 \Rightarrow x^2 - \frac{5}{3}x = -\frac{2}{3}$

Half of $\frac{5}{3}$ is $\frac{5}{6}$, and $\left(\frac{5}{6}\right)^2 = \frac{25}{36}$. Add to both sides: $x^2 - \frac{5}{3}x + \frac{25}{36} = -\frac{2}{3} + \frac{25}{36}$ $\left(x - \frac{5}{6}\right)^2 = \frac{-24 + 25}{36} = \frac{1}{36}$

Taking square roots: $x - \frac{5}{6} = \pm \frac{1}{6}$ $x = \frac{5}{6} + \frac{1}{6} = 1 \quad \text{or} \quad x = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}$

Roots: $x = 1$ and $x = \dfrac{2}{3}$.

Let the number be $x$. Then $x + \dfrac{1}{x} = \dfrac{10}{3}$.

Multiply throughout by $3x$: $3x^2 + 3 = 10x$ $3x^2 - 10x + 3 = 0$

Using the quadratic formula with $a=3, b=-10, c=3$: $b^2 - 4ac = 100 - 36 = 64$ $x = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}$ $x = 3 \quad \text{or} \quad x = \frac{1}{3}$

The number is $3$ (its reciprocal is $\frac{1}{3}$) — both solutions describe the same pair.

Here $a = 2$, $b = 1$, $c = -4$.

Discriminant: $b^2 - 4ac = 1 - 4(2)(-4) = 1 + 32 = 33$.

$x = \frac{-1 \pm \sqrt{33}}{4}$

Since $\sqrt{33} \approx 5.745$: $x = \frac{-1 + 5.745}{4} = \frac{4.745}{4} \approx 1.19$ $x = \frac{-1 - 5.745}{4} = \frac{-6.745}{4} \approx -1.69$

Roots (to two decimals): $x \approx 1.19$ and $x \approx -1.69$.

Let the two consecutive positive integers be $x$ and $x + 1$.

Given: $x(x+1) = 306$, so $x^2 + x - 306 = 0$

Using the quadratic formula with $a=1, b=1, c=-306$: $b^2 - 4ac = 1 + 1224 = 1225$ $x = \frac{-1 \pm \sqrt{1225}}{2} = \frac{-1 \pm 35}{2}$ $x = 17 \quad \text{or} \quad x = -18$

Since the integers are positive, $x = 17$. Hence the integers are $17$ and $18$.

Here $a = 2$, $b = -6$, $c = 5$.

Discriminant: $D = b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4$.

Since $D < 0$, the equation has no real roots (the roots are imaginary/complex).

For two equal real roots, the discriminant must be zero: $b^2 - 4ac = 0$.

Here $a = k$, $b = -2\sqrt{5}$, $c = 4$.

$(-2\sqrt{5})^2 - 4(k)(4) = 0$ $4 \times 5 - 16k = 0$ $20 - 16k = 0$ $k = \frac{20}{16} = \frac{5}{4}$

Hence $k = \dfrac{5}{4}$.

Here $a = 3$, $b = -4\sqrt{3}$, $c = 4$.

Discriminant: $D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.

Since $D = 0$, the equation has two equal real roots.

The repeated root is: $x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$

So both roots equal $\dfrac{2}{\sqrt{3}}$.

For real and equal roots, $b^2 - 4ac = 0$.

Here $a = 1$, $b = k$, $c = 9$.

$k^2 - 4(1)(9) = 0$ $k^2 - 36 = 0$ $k^2 = 36$ $k = \pm 6$

Hence $k = 6$ or $k = -6$.

Here $a = k+1$, $b = -2(k-1)$, $c = 1$.

For equal roots, $b^2 - 4ac = 0$: $[-2(k-1)]^2 - 4(k+1)(1) = 0$ $4(k-1)^2 - 4(k+1) = 0$

Divide by $4$: $(k-1)^2 - (k+1) = 0$ $k^2 - 2k + 1 - k - 1 = 0$ $k^2 - 3k = 0$ $k(k - 3) = 0$ $k = 0 \quad \text{or} \quad k = 3$

(Note: $k = -1$ would make it non-quadratic, so it is excluded, but here neither value is $-1$.) Hence $k = 0$ or $k = 3$.