Quadratic equations with real coefficients

In this section you will study quadratic equations with real coefficients. That means the coefficient so of $x^{2}$,x and constant are real numbers. The coefficients are real but not necessary every time we will get real roots. Some times the roots are complex numbers. The nature of the roots depends on the value of $b^{2}$ - 4ac when the quadratic equation is of the form a$x^{2}$ + bx + c = 0

a$x^{2}$ + bx + c = 0 , a,b,c $\epsilon$ R and a $\neq$ 0.
Multiplying both sides by a , we get
$a^{2}x^{2}$ +abx + ac = 0
$a^{2}x^{2}$ +abx = - ac ( subtract ac on both sides )
$a^{2}x^{2} +abx + \frac{b^{2}}{4} = \frac{b^{2}}{4}$ - ac ( add $\frac{b^{2}}{4}$ on both sides)
$\left ( ax + \frac{b}{2} \right )^{2} = \frac{b^{2}-4ac}{4}$

$\left ( ax + \frac{b}{2} \right) = \pm \frac{\sqrt{b^{2}-4ac}}{2}$

∴ ax = $\frac{- b}{2}\pm \frac{\sqrt{b^{2}-4ac}}{2}$

ax = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2}$

x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
So the roots of quadratic equation $\alpha$ and $\beta$, given by
$\alpha = \frac{-b +\sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

The nature of roots of the quadratic equation is dependent on $b^{2}$-4ac. This is called discriminant and is denoted by 'D'
If $b^{2}$-4ac < 0 then the roots are not real.
And if $b^{2}$-4ac > 0 then the roots are real.
I : If $b^{2}$-4ac = 0 then $\alpha = \beta = \frac{-b}{2a}$. Here the roots are real and equal.
II : If $b^{2}$-4ac is positive and perfect square, then the roots $\alpha$ and $\beta$ are rational and unequal.
III : If $b^{2}$-4ac is positive, then the roots $\alpha$ and $\beta$ are irrational and unequal.
IV : If $b^{2}$-4ac is negative, then the roots $\alpha$ and $\beta$ are imaginary and is given by.
$\alpha = \frac{-b +i\sqrt{4ac - b^{2}}}{2a}$ and $\beta = \frac{-b - i\sqrt{4ac - b^{2}}}{2a}$

Examples on quadratic equations with real coefficients

1) Find the roots of the equation 4$x^{2}$ + 9 = 0
Solution : The given equation 4$x^{2}$ + 9 = 0 is quadratic as its degree is 2 and it has the coefficients which are real.
4$x^{2}$ + 9 = 0
Here, a = 4 , b = 0 and c = 9
$b^{2}$-4ac = $0^{2}-4\times 4 \times$ 9 = -64 < 0 so the roots are imaginary.
$\alpha = \frac{-b +i\sqrt{4ac - b^{2}}}{2a}$ and $\beta = \frac{-b - i\sqrt{4ac - b^{2}}}{2a}$

$\alpha =\frac{0 + i\sqrt{144}}{2\times 4}$ = $\frac{12i}{8} = \frac{3}{2}i$

and $\beta =\frac{0 - i\sqrt{144}}{2\times 4}$ = $\frac{-12i}{8} = \frac{-3}{2}i$

2) Solve the equation $x^{2}$ - 4x + 13 = 0 by factorization method.
Solution : $x^{2}$ - 4x + 13 = 0
$x^{2}$ - 4x + 4 + 9 = 0
$(x - 2)^{2}$ + 9 = 0
$(x - 2)^{2} - 9i^{2}$ = 0
$(x - 2)^{2} - (3i)^{2}$ = 0
[(x - 2) -3i][(x -2) + 3i] = 0
(x - 2 - 3i)(x - 2 + 3i) = 0
∴ x = 2 + 3i or x = 2 - 3i

11th grade math

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