Solve no 4 quadratic equation

1.	Solve no 4 quadratic equation
Answer» Answer : $\bf x= \frac{3+\sqrt{29}}{2} , \frac{3-\sqrt{29}}{2}$ Step-by-step explanation : ➤ Quadratic POLYNOMIALS : ✯ It is a polynomial of degree 2 ✯ General form : ax² + bx + c = 0 ✯ Determinant, D = b² - 4ac ✯ Based on the VALUE of Determinant, we can DEFINE the nature of roots. D > 0 ; real and unequal roots D = 0 ; real and equal roots D < 0 ; no real roots i.e., imaginary ✯ RELATIONSHIP between zeroes and coefficients : ✩ Sum of zeroes = -b/a ✩ Product of zeroes = c/a ________________________________ Given quadratic equation, x² = 3x + 5 x² - 3x - 5 = 0 It is of the form ax² + bx + c = 0 a = 1 , b = -3 , c = -5 By using quadratic formula, $\boxed{\bf x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} }$ Substitute the values, $x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)} }{2(1)} \\\\ x=\frac{3 \pm \sqrt{9+20} }{2} \\\\ x=\frac{3 \pm \sqrt{29} }{2} \\\\\\ \bf \implies x= \frac{3+\sqrt{29}}{2} , \frac{3-\sqrt{29}}{2}$

Answer»

Answer :

$\bf x= \frac{3+\sqrt{29}}{2} , \frac{3-\sqrt{29}}{2}$

Step-by-step explanation :

➤ Quadratic POLYNOMIALS :

✯ It is a polynomial of degree 2

✯ General form :

ax² + bx + c = 0

✯ Determinant, D = b² - 4ac

✯ Based on the VALUE of Determinant, we can DEFINE the nature of roots.

D > 0 ; real and unequal roots

D = 0 ; real and equal roots

D < 0 ; no real roots i.e., imaginary

✯ RELATIONSHIP between zeroes and coefficients :

✩ Sum of zeroes = -b/a

✩ Product of zeroes = c/a

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Given quadratic equation,

x² = 3x + 5

x² - 3x - 5 = 0

It is of the form ax² + bx + c = 0

a = 1 , b = -3 , c = -5

By using quadratic formula,

$\boxed{\bf x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} }$

Substitute the values,

$x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)} }{2(1)} \\\\ x=\frac{3 \pm \sqrt{9+20} }{2} \\\\ x=\frac{3 \pm \sqrt{29} }{2} \\\\\\ \bf \implies x= \frac{3+\sqrt{29}}{2} , \frac{3-\sqrt{29}}{2}$