Question

4.
यदि दिये गये द्विघात समीकरण ax2+bx+cto, a+0 में दो बीज (मूल) के अनुपात 1:r हो तो प्रमाणित
करो- (v+1)2 = b
ac
r​

Answer 1

SOLUTION

TO PROVE

If the roots of the quadratic equation ax² + bx + c = 0 ( a ≠ 0 ) are in the ratio 1 : r then prove that

$\displaystyle \sf{ \frac{ {(r + 1)}^{2} }{r} = \frac{ {b}^{2} }{ac} }$

EVALUATION

Here the given Quadratic equation is

ax² + bx + c = 0 ( a ≠ 0 )

Now it is given that the roots of the quadratic equation are in the ratio 1 : r

Suppose the roots are α and rα

Then we have

$\displaystyle \sf{ \alpha + r \alpha = - \frac{b}{a} }$

$\displaystyle \sf{ \implies \alpha(1 + r) = - \frac{b}{a} }$

$\displaystyle \sf{ \implies \alpha \: = - \frac{b}{a(1 + r)} } \: \: \: - - - (1)$

Again

$\displaystyle \sf{ \alpha \times r \alpha = \frac{c}{a} }$

$\displaystyle \sf{ \implies r{\alpha}^{2} = \frac{c}{a} }$

$\displaystyle \sf{ \implies {\alpha}^{2} = \frac{c}{ar} }$

$\displaystyle \sf{ \implies { \bigg( - \frac{b}{a(1 + r)} \bigg)}^{2} = \frac{c}{ar} }$

$\displaystyle \sf{ \implies \frac{ {b}^{2} }{ {a}^{2} } \bigg[ \frac{1}{{(1 + r) }^{2} } \bigg]= \frac{c}{ar} }$

$\displaystyle \sf{ \implies \frac{r}{{(1 + r) }^{2} } = \frac{ac}{ {b}^{2} } }$

$\displaystyle \sf{ \implies \frac{ {(r + 1)}^{2} }{r} = \frac{ {b}^{2} }{ac} }$

Hence proved

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. If 1/2 is a root of the quadratic equation x²-mx-5/4=0 , then the value of m is?

https://brainly.in/question/21062028

2.Find the roots of the quadratic equation 6x2 +5x + 1 =0 by method of completing the squares.

https://brainly.in/question/30446963

3. Write a quadratic equation whose root are -4 and -5

https://brainly.in/question/24154410