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Factor and use the zero-product property to find the roots of the following quadratic equation.
(a) \[0={{x}^{2}}-7x+12\]
(b) \[0=6{{x}^{2}}-23x+20\]
(c) \[0={{x}^{2}}-9\]
(d) \[0={{x}^{2}}+12x+36\]

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: First understand the definition of zero product property. To factorize the quadratic expression in (a) and (b) apply the middle term split method. Use the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to factorize the expression in (c). For (d) use the formula: - \[{{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}\] for factorization.

Complete step-by-step solution:
Here, we have been provided with four quadratic equations and we are asked to factorize them and use the zero-product property to find the roots. But first we need to know about the zero-product property.
 Now, in mathematics, the zero product property states that if m and n are two non – zero numbers then their product will not be zero. In other words, if \[m\times n=0\] then either m = 0 and n = 0.
Now, let us come to the quadratic equations one – by – one.
(a) \[0={{x}^{2}}-7x+12\]
\[\Rightarrow {{x}^{2}}-7x+12=0\]
Using the middle term split method, we have,
\[\begin{align}
& \Rightarrow {{x}^{2}}-4x-3x+12=0 \\
& \Rightarrow \left( x-4 \right)\left( x-3 \right)=0 \\
\end{align}\]
Applying the zero-product property, we get,
\[\Rightarrow \] either x = 4 or x = 3
Therefore, x = 4 or x = 3 is the solution.
(b) \[0=6{{x}^{2}}-23x+20\]
\[\Rightarrow 6{{x}^{2}}-23x+20=0\]
Using the middle term split method, we get,
\[\begin{align}
& \Rightarrow 6{{x}^{2}}-15x-8x+20=0 \\
& \Rightarrow 3x\left( 2x-5 \right)-4\left( 2x-5 \right)=0 \\
& \Rightarrow \left( 3x-4 \right)\left( 2x-5 \right)=0 \\
\end{align}\]
Applying the zero-product property, we get,
\[\Rightarrow \] either \[x=\dfrac{4}{3}\] or \[x=\dfrac{5}{2}\]
Therefore, \[x=\dfrac{4}{3}\] or \[x=\dfrac{5}{2}\] is the solution.
(c) \[0={{x}^{2}}-9\]
\[\begin{align}
& \Rightarrow {{x}^{2}}-9=0 \\
& \Rightarrow {{x}^{2}}-{{3}^{2}}=0 \\
\end{align}\]
Using the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get,
\[\Rightarrow \] either x = -3 or x = 3
Therefore, x = -3 or x = 3 is the solution.
(d) \[0={{x}^{2}}+12x+36\]
\[\Rightarrow {{x}^{2}}+12x+36=0\]
This can be written as: -
\[\Rightarrow {{x}^{2}}+2\times 6\times x+{{6}^{2}}=0\]
Using the algebraic identity: - \[{{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}\], we get,
\[\begin{align}
& \Rightarrow {{\left( x+6 \right)}^{2}}=0 \\
& \Rightarrow \left( x+6 \right)\left( x+6 \right)=0 \\
\end{align}\]
Applying the zero-product property, we get,
\[\begin{align}
& \Rightarrow x+6=0 \\
& \Rightarrow x=-6 \\
\end{align}\]
Therefore, $x = -6$ is the solution.

Note: One must not apply the discriminant formula to solve the above questions because if you will do so then according to the question it will be considered as a wrong approach. This is because we have to factorize the equations and then use the zero-product property but the discriminant formula will directly give the values of x without factoring. You must remember the zero-product property, middle term split method and all the algebraic identities.

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