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A quadratic equation is a polynomial equation of degree 2 in the variable \( x \). It has the general form:
\[ax^2 + bx + c = 0\]where \( a, b, c \) are real numbers, and \( a \neq 0 \). Here, \( a \) is called the quadratic coefficient, \( b \) the linear coefficient, and \( c \) the constant term.
The solutions (or roots) of the quadratic equation are given by the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The expression under the square root, \( \Delta = b^2 - 4ac \), is called the discriminant. It determines the nature of the roots:
Example: For \( x^2 - 4x + 3 = 0 \), \( a=1, b=-4, c=3 \), so
\[\Delta = (-4)^2 - 4 \times 1 \times 3 = 16 - 12 = 4 > 0\]Thus, two distinct real roots exist.
Let the roots of the quadratic equation \( ax^2 + bx + c = 0 \) be \( \alpha \) and \( \beta \). From the quadratic formula and factorization, the following relations hold:
\[\alpha + \beta = -\frac{b}{a} \quad \text{(Sum of roots)}\]\[\alpha \beta = \frac{c}{a} \quad \text{(Product of roots)}\]These relations are very useful in solving problems without explicitly finding the roots.
The graph of \( y = ax^2 + bx + c \) is a parabola. The roots correspond to the x-intercepts of the parabola.
The parabola opens upwards since \( a = 1 > 0 \). The roots \( x=1 \) and \( x=3 \) are where the parabola crosses the x-axis.
Quadratic equations appear in various contexts such as projectile motion, area problems, optimization, and more. Understanding the nature of roots and their relations helps solve these efficiently.
Solution:
Given \( a=2, b=-7, c=3 \).
Calculate discriminant:
\[ \Delta = (-7)^2 - 4 \times 2 \times 3 = 49 - 24 = 25 \]Since \( \Delta > 0 \), roots are real and distinct.
Apply quadratic formula:
\[ x = \frac{-(-7) \pm \sqrt{25}}{2 \times 2} = \frac{7 \pm 5}{4} \]So,
\[ x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3, \quad x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2} \]Answer: \( x = 3 \) or \( x = \frac{1}{2} \)
Solution:
\[ a=1, \quad b=4, \quad c=5 \] \[ \Delta = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4 < 0 \]Since \( \Delta < 0 \), roots are complex conjugates.
Answer: Roots are complex and not real.
Solution:
Equal roots imply \( \Delta = 0 \).
\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \times 1 \times k = 25 - 4k = 0 \] \[ \Rightarrow 4k = 25 \implies k = \frac{25}{4} = 6.25 \]Answer: \( k = \frac{25}{4} \)
Solution:
Sum of roots:
\[ \alpha + \beta = 3 + (-2) = 1 \]Product of roots:
\[ \alpha \beta = 3 \times (-2) = -6 \]Using the relation \( x^2 - (\alpha + \beta) x + \alpha \beta = 0 \), the equation is:
\[ x^2 - 1x - 6 = 0 \quad \Rightarrow \quad x^2 - x - 6 = 0 \]Answer: \( x^2 - x - 6 = 0 \)
Solution:
We look for two numbers whose product is 8 and sum is 6. These are 4 and 2.
\[ x^2 + 6x + 8 = (x + 4)(x + 2) = 0 \] \[ \Rightarrow x = -4, \quad x = -2 \]Answer: \( x = -4 \) or \( x = -2 \)
Solution:
Let roots be \( 2k \) and \( 3k \).
Sum of roots:
\[ 2k + 3k = 5k = -\frac{p}{3} \]Product of roots:
\[ 2k \times 3k = 6k^2 = \frac{12}{3} = 4 \] \[ \Rightarrow k^2 = \frac{4}{6} = \frac{2}{3} \] \[ k = \sqrt{\frac{2}{3}} \]Sum of roots:
\[ 5k = -\frac{p}{3} \implies p = -15k = -15 \sqrt{\frac{2}{3}} = -15 \times \frac{\sqrt{6}}{3} = -5 \sqrt{6} \]Answer: \( p = -5 \sqrt{6} \)
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