Question

If the sum of the zeros of p(x)=(k-3)x^2-12+5 is 4, then find the value of k.​

Answer 1

Correct question:-

• If the sum of zeroes of p(x) = (k - 3)x² - 12x + 5 is 4, then find the value of k.

Answer:-

Given that, p(x) = (k - 3)x² - 12x + 5, on comparing it with the standard form of quadratic equation, that is ax² + bx + c, we get,

• a = k - 3
• b = -12
• c = 5

We know that, for a polynomial ax² + bx + c, having roots α and β,

• sum of roots = α + β = - (coefficient of x)/(coefficient of x²) = -b/a
• product of roots = αβ = (constant term)/(coefficient of x²) = c/a

Let the roots of p(x) = (k - 3)x² - 12x + 5 be α and β. Using the relation of sum of roots, we get,

α + β = -b/a

→ α + β = -(-12)/(k - 3)

→ α + β = 12/(k - 3)

Now, it is given that, the value of the sum of roots is 4. So,

12/(k - 3) = 4

→ 12 = 4(k - 3)

→ 12 = 4k - 12

→ 4k = 24

→ k = 6 Ans.

Answer 2

Question :-

• If the sum of the zeros of p(x) = (k-3)x² - 12x + 5 is 4 then Find the value of k.

Given :-

• Sum of zeros = 4

To FinD :-

• The value of k.

Solution :-

Comparing the given polynomial by ax² + bx + c = 0

• a = (k - 3)

• b = - 12

• c = 5

We know that :-

• $\sf \: sum \: of \: zeros = - \frac{b}{a}$

So,

$\star \sf \: sum \: of \: zeros = - \frac{( - 12)}{(k - 3)} \\ \\ \mapsto \sf \: 4 = \frac{12}{k - 3} \\ \\ \mapsto \sf \: 4(k - 3) = 12 \\ \\ \mapsto \sf \: 4k - 12 = 12 \\ \\ \mapsto \sf \: 4k = 12 + 12 \\ \\ \mapsto \sf \: 4k = 24 \\ \\ \mapsto \sf \: k = \frac{24}{4} \\ \\ \mapsto \boxed{ \sf \: k = 6}$

Hence, the value of k is 6.

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