Question

If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1)x -1, then find the value of a.​

Answer 1

Solution:

Given:

→ p(x) = ax² - 3(a - 1)x - 1

→ 1 is a zero of p(x).

Therefore,

→ p(1) = 0

Putting x = 1 in p(x), we get,

→ a × (1)² - 3 × (a - 1) × 1 - 1 = 0

→ a - 3(a - 1) - 1 = 0

→ a - 3a + 3 - 1 = 0

→ -2a + 2 = 0

→ -2a = -2

Dividing both sides by -2, we get,

→ a = 1

★ So, the value of a is 1.

Answer:

• a = 1

Verification:

Put a = 1 in p(x), the polynomial becomes -

→ p(x) = x² - 3 × (1 - 1) × x - 1

→ p(x) = x² - 1

Now, put x = 1. We get,

→ p(1) = (1)² - 1

→ p(1) = 0

As p(1) = 0, 1 is a zero of p(x).

So, our answer is correct (Verified)

Answer 2

Answer:

a=1

Step-by-step explanation:

Given, 1 is a zero of polynomial P(x)

∴$P(1)=0$

$P(x)= ax^{2} -3(a-1)x-1$

$P(1)=0=a(1)^{2} -3(a-1)(1)-1\\a-3a +3-1=0\\-2a+2=0\\2a=2\\a=1$

Hope it helps...

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