# A quadratic equation whose one root is 3 is

(a) ${{z}^{2}}-6z-5=0$

(b) ${{z}^{2}}+6z-5=0$

(c) ${{z}^{2}}-5z-6=0$

(d) ${{z}^{2}}-5z+6=0$

Answer

Verified

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Hint: The only way to solve this question is through option verification. It is given that the root is 3. This means, if we substitute z = 3 in the quadratic equations, the left-hand side must be equal to the right-hand side. We will consider each option one by one and substitute z = 3. If the left-hand side is equal to the right-hand side, that option is the correct option.

Let us first consider the option (a). We will substitute z = 3 to see whether the lefthand side is equal to the right-hand side or not.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-6\left( 3 \right)-5$

$\Rightarrow $ LHS = 9 – 18 – 5

$\Rightarrow $ LHS = ─24

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (a) does not verify.

Now, let us consider option (b) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}+6\left( 3 \right)-5$

$\Rightarrow $ LHS = 9 + 18 – 5

$\Rightarrow $ LHS = 22

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (b) does not verify.

Now, we will consider option (c) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)-6$

$\Rightarrow $ LHS = 9 – 15 – 6

$\Rightarrow $ LHS = ─12

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (c) does not verify.

The option left with us is an option (d) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)+6$

$\Rightarrow $ LHS = 9 – 15 + 6

$\Rightarrow $ LHS = 0

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS = RHS

Thus, option (d) verify with the conditions.

**Complete step-by-step solution:**Let us first consider the option (a). We will substitute z = 3 to see whether the lefthand side is equal to the right-hand side or not.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-6\left( 3 \right)-5$

$\Rightarrow $ LHS = 9 – 18 – 5

$\Rightarrow $ LHS = ─24

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (a) does not verify.

Now, let us consider option (b) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}+6\left( 3 \right)-5$

$\Rightarrow $ LHS = 9 + 18 – 5

$\Rightarrow $ LHS = 22

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (b) does not verify.

Now, we will consider option (c) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)-6$

$\Rightarrow $ LHS = 9 – 15 – 6

$\Rightarrow $ LHS = ─12

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS $\ne $ RHS

Thus, option (c) does not verify.

The option left with us is an option (d) and substitute z = 3.

$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)+6$

$\Rightarrow $ LHS = 9 – 15 + 6

$\Rightarrow $ LHS = 0

$\Rightarrow $ RHS = 0

$\Rightarrow $ LHS = RHS

Thus, option (d) verify with the conditions.

**Hence, option (d) is the correct option.****Note:**The other method is to actually solve the equations and find the roots of that equation. Those equations can be solved with the factorization method, completing the square method or formula method. But if it is possible to verify the equations, we shall opt for verification as it is quicker and relatively simple.Last updated date: 25th Sep 2023

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