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Let us put this to practice. Example 1: Discuss the nature of the roots of the quadratic equation 2x 2 – 8x + 3 = 0. Solution: Here the coefficients are all rational. The discriminant D of the given equation is. D = b 2 – 4ac = (-8) 2 – 4 x 2 x 3. = 64 – 24. = 40 > 0.
Finding the quadratic equation from its given roots. 0. A new method to derive the quadratic formula. 3.
Jul 7, 2019 · Closed 5 years ago. Our maths teacher taught us that conditions for roots of a quadratic equation ax2 + bx + c = 0 a x 2 + b x + c = 0 to lie at infinity are : A) (for exactly one root at infinity) a = 0 a = 0, b b = non-zero, & c c can be zero or non-zero. B) (for both roots at infinity) a = 0 a = 0, b = 0 b = 0, and c c = non-zero.
Aug 19, 2013 · Yes. All you have to do is write α = − 4 − √21, β = − 4 + √21, and the desired quadratic equation is (x − α / β)(x − β / α). That's it. The most important thing is to note that if you are asked to find the quadratic equation with roots a and b, the answer is just (x − a)(x − b). You don't have to look for relationships ...
If the A.M. and G.M. of roots of a quadratic equations are 8 and 5 respectively, then the quadratic equation will be
If − 5 is a root of the quadratic equation 2 x 2 + p x − 15 = 0 and the quadratic equation p (x 2 + x) + k = 0 has equal roots, find the value of k. View Solution Q 3
If roots is the equation (a–b)x 2 +(b–c)x+(c–a)=0 are areare equal then prove that 2a=b+c.. View Solution. Q5
Jan 27, 2023 · 1. The formula your friend gave you is a special case of Vieta's formulas. So, using this formula, we find that the quadratic equation is x2 − (−4 + 0.4i + −4 − 0.4i)x + (−4 + 0.4i)(−4 − 0.4i) =x2 + 8x + 16.16 x 2 − (− 4 + 0.4 i + − 4 − 0.4 i) x + (− 4 + 0.4 i) (− 4 − 0.4 i) = x 2 + 8 x + 16.16. Multiply this ...
Jul 4, 2016 · Let $ax^2+bx+c=0$ be the quadratic. I usually start by looking if: 1) $a+b+c=0$ then the roots are $1$ and $c/a$.
Here is a hint to develop your existing method. Multiply the equation in a and b by 2 and add a constant which enables you to factorise it. Given x2 + 2kx + k = 0 ⇒ x2 + 2kx + k2 = k2 − k ⇒ (x + k)2 = k2 − k. So we get (x + k)2 = (√k2 − k)2 ⇒ x + k = ± √k2 − k. So we get x = ± √k2 − k − k, Now here x ∈ Z. So let k2 ...
