Question

find the zeros of each of the following polynomials and verify the relationship between the zeros and the coefficients 7 x square - 25 x - 12 CBSE 10th class

Answer 1

Answer:

$\boxed{\green{\alpha = \frac{4}{7} \: & \: \beta = 3}}$

Step-by-step explanation:

Here we are Given

$p(x) = 7x^{2} - 25x - 12$

What to do here?

• Find zeroes of given polynomial [p(x)]
• State and verify relation between Zeroes and co-efficients

Solution

Calculating Zeroes

${Let, \: here} \\\\\: \: \: p(x) = 0\\\implies 7x^{2} - 25x +12 = 0\\\\\\ Factorizing this equation we get \\\implies 7x^{2} - 25x +12 = 0 \\\implies 7x^{2} - (21 + 4)x +12 = 0\\\implies 7x^{2} - 21x - 4x + 12 = 0\\\implies 7x(x - 3) -4(x-3) = 0\\\implies (7x-4)(x-3) = 0\\ Now here we observe that if p(x) = 0 then\\either of the following conditions would be possible\\ \\x= \frac{4}{7} \\or \: x = 3\\\\ \therefore We can say zeroes of p(x) are \: \frac{4}{7} \: and \: 3$

Verifying Relationship between zeroes and coefficients

$\because every quadratic equation is in form \\ax^{2} + bx + c = 0\\\\ So, let \frac{c}{a} = LHS\\\frac{12}{7} = LHS \\& RHS = \alpha \beta \\\therefore RHS = \frac{4}{7} \times 3 \\\implies RHS = \frac{12}{7}\\\\ Thus we can say LHS = RHS \\ Similarly we can Prove \alpha +\beta = \frac{-b}{a}$

Hope I Made it clear! :)

Happy Learning !